MCQ 11 Mark
The aqueous solutions of sodium formate, anilinium chloride and potassium cyanide are respectively.
Answerbasic, acidic, basicExplanation:
$\begin{gathered}\underset{\text { Basic in nature }}{ HCOONa }+ H - OH \rightleftharpoons \underset{\text { strong base }}{ NaOH }+ H - COOH \\ C _6 H _5 NH _3 Cl ^{-}+ H - OH \rightleftharpoons \underset{\text { acidic }}{ H _3 O ^{+}}+ C _6 H _5- NH _2+ Cl ^{-} \\ \underset{\text { basic }}{ KCN }+ H - OH \rightleftharpoons \underset{\text { strong base }}{ KOH }+\underset{\text { weak acid }}{ HCN }\end{gathered}$
View full question & answer→MCQ 21 Mark
Which of these is not likely to act as Lewis base?
AnswerCorrect option: A. $BF_3$
$BF_3$
Explanation:
$BF_3$ → electron deficient → Lewis acid
$PF_3$ → electron rich → Lewis base
CO → having lone pair of electron → Lewis base
$F^-$ → unshared pair of electron → lewis base
View full question & answer→MCQ 31 Mark
Which of the following fluro compounds is most likely to behave as a Lewis base?
- A
$BF_3$
- ✓
$PF_3$
- C
$CF_4$
- D
$SiF_4$
AnswerCorrect option: B. $PF_3$
$PF _3$ Explanation:
$BF _3 \rightarrow$ electron deficient $\rightarrow$ Lewis acid
$PF _3 \rightarrow$ electron rich $\rightarrow$ Lewis base
$CF _4 \rightarrow$ neutral $\rightarrow$ neither lewis acid nor base
$SiF _4 \rightarrow$ neutral $\rightarrow$ neither lewis acid nor base
View full question & answer→MCQ 41 Mark
Which will make basic buffer?
- A
50 mL of 0.1 M NaOH + 25 mL of 0.1 M $CH_3COOH$
- B
100 mL of 0.1 M $CH_3COOH + 100$ mL of 0.1 M $NH_4OH$
- ✓
100 mL of 0.1 M HCI + 200 mL of 0.1 M $NH_4OH$
- D
100 mL of 0.1 M HCI + 100 mL of 0.1 M NaOH
AnswerCorrect option: C. 100 mL of 0.1 M HCI + 200 mL of 0.1 M $NH_4OH$
100 mL of 0.1 M HCI + 200 mL of 0.1 M $NH_4OH$
Explanation:
Basic buffer is the solution which has weak base and its salt
$\underset{200 ml }{ NH _4 OH }+\underset{100 ml }{ HCl } \longrightarrow \underset{\text { Salt }}{ NH _4 Cl }+ H _2 O +\underset{(100 ml \text { weak base })}{ NH _4 OH }$
View full question & answer→MCQ 51 Mark
Conjugate base for Bronsted acids $\mathrm{H_2O}$ and HF are ___________.
- A
$\mathrm{OH}^{-}$and $\mathrm{H}_2 \mathrm{FH}^{+}$, respectively
- B
$\mathrm{H}_3 \mathrm{O}^{+}$and $\mathrm{F}^{-}$, respectively
- ✓
$\mathrm{OH}^{-}$and $\mathrm{F}^{-}$, respectively
- D
$\mathrm{H}_3 \mathrm{O}^{+}$and $\mathrm{H}_2 \mathrm{~F}^{+}$, respectively
AnswerCorrect option: C. $\mathrm{OH}^{-}$and $\mathrm{F}^{-}$, respectively
Conjugate base for Bronsted acids $\mathrm{H_2O}$ and HF are $\mathrm{OH}^{-}$and $\mathrm{F}^{-}$, respectively
Explanation:
$\begin{aligned} & \underset{\text { acid 1 }}{ H _2 O }+\underset{\text { base 1 }}{ H _2 O } \rightleftharpoons \underset{\text { acid 2 }}{ H _3 O ^{+}}+\underset{\text { base 2 }}{ OH ^{-}} \\ & \underset{\text { acid 1 }}{ HF }+\underset{\text { base 1 }}{ H _2 O } \rightleftharpoons \underset{\text { acid 2 }}{ H _3 O ^{+}}+\underset{\text { base 2 }}{ F ^{-}}\end{aligned}$
∴ Conjugate bases are $\mathrm{OH}^{-}$and $\mathrm{F}^{-}$, respectively
View full question & answer→MCQ 61 Mark
pH of a saturated solution of $Ca(OH)_2$ is 9. The Solubility product $(K_{sp})$ of $Ca(OH)_2$
AnswerCorrect option: A. $ 0.5 \times 10^{-15} $
$0.5 \times 10^{-15}$
Explanation:
$
Ca ( OH )_2 \rightleftharpoons Ca ^{2+}+2 OH ^{-}
$
Given that $pH =9$
$
\begin{aligned}
& pOH =14-9=5 \\
& {\left[ pOH =-\log _{10}\left[ OH ^{-}\right]\right]} \\
& \therefore\left[ OH ^{-}\right]=10^{- pOH } \\
& {\left[ OH ^{-}\right]=10^{-5} M } \\
& K _{ sp }=\left[ Ca ^{2+}\right]\left[ OH ^{-}\right]^2 \\
& =\frac{10^{-5}}{2} \times\left(10^{-5}\right)^2 \\
& =0.5 \times 10^{-15}
\end{aligned}
$
View full question & answer→MCQ 71 Mark
The solubility of $\mathrm{BaSO}_4$ in water is $2.42 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}$ at 298 K . The value of its solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$ will be: (Given molar mass of $\mathrm{BaSO}_4=233 \mathrm{~g} \mathrm{~mol}^{-1}$ )
- A
$ 1.08 \times 10^{-14} \mathrm{~mol}^2 \mathrm{~L}^{-2} $
- B
$ 1.08 \times 10^{-12} \mathrm{~mol}^2 \mathrm{~L}^{-2} $
- ✓
$ 1.08 \times 10^{-10} \mathrm{~mol}^2 \mathrm{~L}^{-2} $
- D
$ 1.08 \times 10^{-8} \mathrm{~mol}^2 \mathrm{~L}^{-2} $
AnswerCorrect option: C. $ 1.08 \times 10^{-10} \mathrm{~mol}^2 \mathrm{~L}^{-2} $
$1.08 \times 10^{-10} mol ^2 L ^{-2}$
Explanation:
$
BaSO _4 \rightleftharpoons Ba ^{2+}+ SO _4^{2-}
$
$
\begin{aligned}
& K_{s p}=(s)(s) \\
& K_{s p}=(s)^2
\end{aligned}
$
$
K_{s p}=(s)^2
$
$
=\left(2.42 \times 10^{-3} g L ^{-1}\right)^2
$
$=\left(\frac{2.42 \times 10^{-3} g L ^{-1}}{233 g mol ^{-1}}\right)^2$
$=\left(0.01038 \times 10^{-3}\right)^2$
$
=\left(1.038 \times 10^{-5}\right)^2
$
$
=1.077 \times 10^{-10}
$
$
=1.08 \times 10^{-10} mol ^2 L ^{-2}
$
View full question & answer→MCQ 81 Mark
Following solutions were prepared by mixing different volumes of $NaOH$ of $HCl$ different concentrations.
i. $60 mL \frac{ M }{10} HCl +40 mL \frac{ M }{10} NaOH$
ii. $55 mL \frac{ M }{10} HCl +45 mL \frac{ M }{10} NaOH$
iii. $75 mL \frac{ M }{5} HCl +25 mL \frac{ M }{5} NaOH$
iv. $100 mL \frac{ M }{10} HCl +100 mL \frac{ M }{10} NaOH$
$pH$ of which one of them will be equal to 1 ?
AnswerExplanation:
$
75 mL \frac{ M }{5} HCl +25 mL \frac{ M }{5} NaOH
$
No of moles of $HCl =0.2 \times 75 \times 10^{-3}=15 \times 10^{-3}$
No of moles of $NaOH =0.2 \times 25 \times 10^{-3}=5 \times 10^{-3}$
No of moles of $HCl$ after mixing $=15 \times 10^{-3}-5 \times 10^{-3}=10 \times 10^{-3}$
$
\therefore \text { Concentration of } HCl =\frac{\text { No of moles of } HCl }{ Vol \text { in litre }}
$
$
\begin{aligned}
& =\frac{10 \times 10^{-3}}{100 \times 10^{-3}} \\
& =0.1 M
\end{aligned}
$
for (iii) solution, $pH$ of $0.1 M HCl =-\log _{10}(0.1)=1$
View full question & answer→MCQ 91 Mark
Dissociation constant of $\mathrm{NH}_4 \mathrm{OH}$ is $1.8 \times 10^{-5}$ the hydrolysis constant of $\mathrm{NH}_4 \mathrm{Cl}$ would be ____________.
- A
$ 1.8 \times 10^{-19} $
- ✓
$ 5.55 \times 10^{-10} $
- C
$ 5.55 \times 10^{-5} $
- D
$ 1.80 \times 10^{-5} $
AnswerCorrect option: B. $ 5.55 \times 10^{-10} $
Dissociation constant of $NH _4 OH$ is $1.8 \times 10^{-5}$ the hydrolysis constant of $NH _4 Cl$ would be $\underline{5.55 \times 10^{-10}}$.
Explanation:
$
\begin{aligned}
& K_h=\frac{K_w}{K_b}=\frac{1 \times 10^{-14}}{1.8 \times 10^{-5}} \\
& =0.55 \times 10^{-9} \\
& =5.5 \times 10^{-10}
\end{aligned}
$
View full question & answer→Question 101 Mark
Which of the following relation is correct for degree of hydrolysis of ammonium acetate?
$
\begin{aligned}
& h=\sqrt{\frac{ K _{ h }}{ C }} \\
& h =\sqrt{\frac{ K _{ a }}{ K _{ b }}} \\
& h =\sqrt{\frac{ K _{ h }}{ K _{ a } \cdot K _{ b }}} \\
& h =\sqrt{\frac{ K _{ a } \cdot K _{ b }}{ K _{ h }}}
\end{aligned}
$
Answer$h=\sqrt{\frac{K_h}{K_a \cdot K_b}}$
View full question & answer→Question 111 Mark
The hydrogen ion concentration of a buffer solution consisting of a weak acid and its salts is given by
$
\left[ H ^{+}\right]= K _{ a } \frac{[ acid ]}{[\text { salt }]}
$
$
\left.\left[ H ^{+}\right]= K _{ a } \text { [salt }\right]
$
$
\left[ H ^{+}\right]= K _{ a } \text { [acid] }
$
$
\left[ H ^{+}\right]= K _{ a } \frac{[\text { salt }]}{[\text { acid }]}
$
AnswerThe hydrogen ion concentration of a buffer solution consisting of a weak acid and its salts is given by $\left[ H ^{+}\right]= K _{ a } \frac{[\text { acid }]}{[\text { salt }]}$.
Explanation:
According to Henderson equation
$
\begin{aligned}
& pH = pK _{ a }+\log \frac{[\text { salt }]}{[\text { acid }]} \\
& \text { i.e., }-\log \left[ H ^{+}\right]=-\log K _{ a }+\log \frac{[\text { salt }]}{[\text { acid }]} \\
& -\log \left[ H ^{+}\right]=\log \frac{[\text { salt }]}{[\text { acid }]} \times \frac{1}{ K _{ a }} \\
& \log \frac{1}{\left[ H ^{+}\right]}=\log \frac{[\text { salt }]}{[\text { acid }]} \times \frac{1}{ K _{ a }} \\
& \therefore\left[ H ^{+}\right]= K _{ a } \frac{[\text { acid }]}{[\text { salt }]}
\end{aligned}
$
View full question & answer→MCQ 121 Mark
The pH of an aqueous solution is Zero. The solution is ____________.
AnswerThe pH of an aqueous solution is Zero. The solution is strongly acidic.Explanation:
$ \mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right] $
$ \therefore\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}} $
$ =10^0 $
$ =1 $
$ {\left[\mathrm{H}^{+}\right]=1 \mathrm{M}} $
The, solution is strongly acidic.
View full question & answer→MCQ 131 Mark
Concentration of the $\mathrm{Ag}^{+}$ ions in a saturated solution of $\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4$ is $2.24 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$ solubility product of $\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4$ is $........$
- A
$ 2.24 \times 10^{-8} \mathrm{~mol}^3 \mathrm{~L}^{-3} $
- ✓
$ 2.66 \times 10^{-12} \mathrm{~mol}^3 \mathrm{~L}^{-3} $
- C
$ 4.5 \times 10^{-11} \mathrm{~mol}^3 \mathrm{~L}^{-3} $
- D
$ 4.5 \times 10^{-11} \mathrm{~mol}^3 \mathrm{~L}^{-3} $
AnswerCorrect option: B. $ 2.66 \times 10^{-12} \mathrm{~mol}^3 \mathrm{~L}^{-3} $
$ 2.66 \times 10^{-12} \mathrm{~mol}^3 \mathrm{~L}^{-3} $
View full question & answer→MCQ 141 Mark
Which of the following can act as Lowry – Bronsted acid as well as base?
- A
- B
$SO _4^{2-}$
- ✓
$HPO _4^{2-}$
- D
AnswerCorrect option: C. $HPO _4^{2-}$
$HPO _4^{2-}$
Explanation:
$HPO _4^{2-}$ can have the ability to accept a proton to form $H _2 PO _4^{-}$
It can also have the ability to donate a proton to form $PO _4^{-3}$
View full question & answer→MCQ 151 Mark
$H _2 PO _4^{-}$ the conjugate base of ____________.
- A
$PO _4^{3-}$
- B
$P_2O-5$
- ✓
$H_3PO_4$
- D
$HPO _4^{2-}$
AnswerCorrect option: C. $H_3PO_4$
$H _2 PO _4^{-}$the conjugate base of $\underline{H}_3 PO _4$.
Explanation:
$\underset{\text { acid 1 }}{ H _3 PO _4}+\underset{\text { base 1 }}{ H }- OH \rightleftharpoons \underset{\text { acid 2 }}{ H _3 O ^{+}}+\underset{\text { base }}{ H _2 PO _4^{-}}$
$\therefore H _2 PO _4^{-}$is the conjugate base of $H _3 PO _4$
View full question & answer→MCQ 161 Mark
The pH of $10^{-5}$ M KOH solution will be ____________.
AnswerThe pH of $10^{-5}$ M KOH solution will be 9.
Explanation:
$\underset{10^{-5} m }{ KOH } \longrightarrow \underset{10^{-5} m }{ K ^{+}}+\underset{10^{-5} m }{ OH ^{-}}$
$[OH^-] = 10^{-5}M$
$pH = 14 – pOH$
$pH = 14 – (– log [OH^-])$
$= 14 + log [OH^-]$
$= 14 + log 10^{-5}$
$= 14 – 5$
$= 9$
View full question & answer→MCQ 171 Mark
The dissociation constant of a weak acid is $1 \times 10^{-3}$. In order to prepare a buffer solution with a pH $=$ 4 , the $\frac{[ Acid ]}{[ Salt ]}$ ratio should be
- A
$4: 3$
- B
$3: 4$
- C
$10: 1$
- ✓
$1: 10$
AnswerCorrect option: D. $1: 10$
The dissociation constant of a weak acid is $1 \times 10^{-3}$. In order to prepare a buffer solution with a pH $=$ 4 , the $\frac{[\text { Acid }]}{[\text { Salt }]}$ ratio should be $\underline{1: 10}$.
Explanation:
$
\begin{aligned}
& K _{ a }=1 \times 10^{-3} \\
& pH =4 \\
& \frac{[\text { Salt }]}{[\text { Acid }]}=? \\
& pH = pK _{ a }+\log \frac{[\text { Salt }]}{[\text { Acid }]} \\
& 4=-\log _{10}\left(1 \times 10^{-3}\right)+\log \frac{[\text { Salt }]}{[\text { Acid }]} \\
& 4=3+\log _{10} \frac{[\text { Salt }]}{[\text { Acid }]} \\
& 1=\log { }_{10} \frac{[\text { Salt }]}{[\text { Acid }]} \\
& \frac{[\text { Salt }]}{[\text { Acid }]}=10^1 \\
& \text { i.e., } \frac{[\text { Acid }]}{[\text { Salt }]}=\frac{1}{10}=1: 10
\end{aligned}
$
View full question & answer→MCQ 181 Mark
What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?
Answer12.65
Explanation:
$
x ml \text { of } 0.1 m NaOH +x ml \text { of } 0.01 M HCl
$
No. of moles of $NaOH =0.1 \times x \times 10^{-3}=0.1 x \times 10^{-3}$
No. of moles of $HCl =0.01 \times x \times 10^{-3}=0.01 \times \times 10^{-3}$
No. of moles of $NaOH$ after mixing $=0.1 x \times 10^{-3}-0.01 x \times 10^{-3}$
$
=0.09 \times \times 10^{-3}
$
$
\text { Concentration of } NaOH =\frac{0.09 x \times 10^{-3}}{2 x \times 10^{-3}}=0.045
$
$
\left[ OH ^{-}\right]=0.045
$
$
\begin{aligned}
& p ^{ OH }=-\log \left(4.5 \times 10^{-2}\right) \\
& =2-\log 4.5 \\
& =2-0.65
\end{aligned}
$
$
=2-0.65
$
$
\begin{aligned}
& =1.35 \\
& pH =14-1.35=12.65
\end{aligned}
$
View full question & answer→MCQ 191 Mark
MY and $NY_3$, are insoluble salts and have the same $K_{sp}$ values of $6.2 × 10^{-13}$ at room temperature. Which statement would be true with regard to MY and NY?
AnswerCorrect option: D. The molar solubility of MY in water is less than that of $NY_3$
The molar solubility of MY in water is less than that of $NY _3$
Explanation:
Addition of salt $KY$ (having a common ion $Y ^{-}$) decreases the solubility of $MY$ and $NY _3$ due to common ion effect.
Option (a) and (b) are wrong.
For salt MY,
$
MY \rightleftharpoons M ^{+}+ Y ^{-}
$
$K_{s p}=(s)(s)$
$
6.2 \times 10^{-13}=s^2
$
$
\therefore s=\sqrt{6.2 \times 10^{-13}} \simeq 10^{-7}
$
for salt $NY _3$,
$
NY _3 \rightleftharpoons N ^{3+}+3 Y ^{-}
$
$
\begin{aligned}
& K_{s p}=(s)(3 s)^3 \\
& K_{s p}=27 s^4
\end{aligned}
$
$
\begin{aligned}
& s=\left(\frac{6.2 \times 10^{-13}}{27}\right)^{\frac{1}{4}} \\
& s \simeq 10^{-4}
\end{aligned}
$
The molar solubility of $MY$ in water is less than of $NY _3$
View full question & answer→MCQ 201 Mark
If the solubility product of lead iodide is $3.2 × 10^{-8}$, its solubility will be ____________.
- ✓
$ 2 \times 10^{-3} \mathrm{M} $
- B
$ 4 \times 10^{-4} \mathrm{M} $
- C
$ 1.6 \times 10^{-5} \mathrm{M} $
- D
$ 1.8 \times 10^{-5} \mathrm{M} $
AnswerCorrect option: A. $ 2 \times 10^{-3} \mathrm{M} $
If the solubility product of lead iodide is $3.2 \times 10^{-8}$, its solubility will be $\underline{2 \times 10^{-3} M }$.
Explanation:
$
PbI _{2( s )} \rightleftharpoons Pb _{( aq )}^{2+}+2 I _{( aq )}^{-}
$
$
K _{ sp }=( s )(2 s )^2
$
$
3.2 \times 10^{-8}=4 s^3
$
$
\begin{aligned}
& S=\left(\frac{3.2 \times 10^{-8}}{4}\right)^{\frac{1}{3}} \\
& =\left(8 \times 10^{-9}\right)^{\frac{1}{3}} \\
& =2 \times 10^{-3} M
\end{aligned}
$
View full question & answer→MCQ 211 Mark
The solubility of $AgCl_{(s)}$ with solubility product $1.6 × 10^{-10}$ in 0.1 M NaCl solution would be ____________.
- A
$ 1.26 \times 10^{-5} \mathrm{M} $
- ✓
$ 1.6 \times 10^{-9} \mathrm{M} $
- C
$ 1.6 \times 10^{-11} \mathrm{M} $
- D
AnswerCorrect option: B. $ 1.6 \times 10^{-9} \mathrm{M} $
The solubility of $AgCl _{( s )}$ with solubility product $1.6 \times 10^{-10}$ in $0.1 M NaCl$ solution would be $\underline{ 1 . 6 \times}$ $10^{-9} M$.
Explanation:
$
\begin{gathered}
AgCl _{( s )} \rightleftharpoons Ag _{( aq )}^{+}+ Cl _{( aq )}^{-} \\
\underset{0.1 M }{ NaCl } \longrightarrow \underset{0.1 M }{ Na ^{+}}+\underset{0.1 M }{ Cl ^{-}}
\end{gathered}
$
$
\begin{aligned}
& K _{ sp }=1.6 \times 10^{-10} \\
& K _{ sp }=\left[ Ag ^{+}\right]\left[ Cl ^{-}\right] \\
& K _{ sp }=( S )( S +0.1)
\end{aligned}
$
$
\begin{aligned}
& 0.1>>>S \\
& \therefore S+0.1 \simeq 0.1 \\
& \therefore S=\frac{1.6 \times 10^{-10}}{0.1} \\
& =1.6 \times 10^{-9}
\end{aligned}
$
View full question & answer→MCQ 221 Mark
Equal volumes of three acid solutions of pH 1, 2 and 3 are mixed in a vessel. What will be the $H^+$ ion concentration in the mixture?
- ✓
$3.7 × 10^{-2}$
- B
$10^{-6}$
- C
- D
AnswerCorrect option: A. $3.7 × 10^{-2}$
$3.7 \times 10^{-2}$
Explanation:
$
\begin{aligned}
& pH =-\log _{10}\left[ H ^{+}\right] \\
& \therefore\left[ H ^{+}\right]=10^{- pH }
\end{aligned}
$
Let the volume be $x mL$
$
V_1 M_1+V_2 M_2+V_3 M_3=V M
$
$\therefore x mL$ of $10^{-1} M + x mL$ of $10^{-2} M + x mL$ of $10^{-3} M =3 x mL$ of $\left[ H ^{+}\right]$
$\therefore\left[ H ^{+}\right]=\frac{ x [0.1+0.01+0.001]}{3 x }$
$
=\frac{0.1+0.01+0.001}{3}
$
$
=\frac{0.111}{3}
$
$
=0.037
$
$
=3.7 \times 10^{-2}
$
View full question & answer→MCQ 231 Mark
The percentage of pyridine $\left( C _5 H _5 N\right)$ that forms pyridinium ion $\left( C _5 H _5 NH \right)$ in a 0.10 M aqueous pyridine solution $\left( K _b\right.$ for $C _5 H _5 N=1.7 \times 10^{-9}$ ) is ............
AnswerThe percentage of pyridine $\left( C _5 H _5 N \right)$ that forms pyridinium ion $\left( C _5 H _5 NH \right)$ in a $0.10 M$ aqueous pyridine solution $\left( K _{ b }\right.$ for $\left.C _5 H _5 N =1.7 \times 10^{-9}\right)$ is $\underline{ 0 . 0 1 3 \% }$.
Explanation:
$
C _5 H _5 N + H - OH \rightleftharpoons C _5 H _5 \stackrel{+}{ N } H + OH ^{-}
$
$
\begin{aligned}
& \frac{\alpha^2 C }{1-\alpha}=K_b \\
& \alpha^2 C \approx K_b \\
& \alpha=\frac{\sqrt{K_b}}{C}=\frac{\sqrt{1.7 \times 10^{-9}}}{0.1} \\
& =\sqrt{1.7} \times 10^{-4} \\
& \text { Percentage of dissociation }=\sqrt{1.7} \times 10^{-4} \times 100 \\
& =1.3 \times 10^{-2} \\
& =0.013 \%
\end{aligned}
$
View full question & answer→MCQ 241 Mark
Sodium chloride is purified by passing HCl gas in a impure solution of sodium chloride. It is based on ………………
MCQ 251 Mark
pH of water is 7. When a substance Y is added in water, the pH becomes 13. The substance Y is a salt of …………..
- A
strong acid and strong base
- B
- C
strong acid and weak base
- ✓
weak acid and strong base
AnswerCorrect option: D. weak acid and strong base
weak acid and strong base
View full question & answer→MCQ 261 Mark
Ionic product of water increases if ………….
- A
- B
$H^+$ is added
- C
$OH^-$ is added
- ✓
Answer(d) temperature increases
Solution:
$K_w$ increases with increase in temperature.
View full question & answer→MCQ 271 Mark
The sotubility of AgCI will be minimum in ………….
- A
$0.001 M AgNO _3$
- B
- ✓
$0.01 M CaCl _2$
- D
$0.01 M NaCl$
AnswerCorrect option: C. $0.01 M CaCl _2$
$0.01 M CaCl _2$
$0.01 M CaCl _2$ gives maximum $Cl ^{-}$ions to keep $K _{ sp }$ of $AgCl$ constant, decrease in $\left[ Ag ^{+}\right]$will be maximum.
View full question & answer→MCQ 281 Mark
Which pair will show common ion effect?
AnswerCorrect option: C. $NH _4 OH + NH _4 Cl$
$NH _4 OH + NH _4 Cl$
View full question & answer→MCQ 291 Mark
What is the $pH$ value of $\frac{N}{100} KOH$ solution?
Answer(d) 11
$10^{-3} N KOH$ will give $\left[ OH ^{-}\right]=10^{-3} M$
$pOH =3$
$pH + pOH =14$
$pH =14-3=11$
View full question & answer→MCQ 301 Mark
What is the correct representation of the solubility product constant of $Ag _2 CrO _4$ ?
- ✓
$\left[ Ag ^{+}\right]^2\left[ CrO _4^{-2}\right]$
- B
$\left[ Ag ^{+}\right]\left[ CrO _4^{-2}\right]$
- C
$\left[2 Ag ^{+}\right]\left[ CrO _4^{-2}\right]$
- D
$\left[2 Ag ^{+}\right]^2\left[ CrO _4^{-2}\right]$
AnswerCorrect option: A. $\left[ Ag ^{+}\right]^2\left[ CrO _4^{-2}\right]$
(a) $\left[ Ag ^{+}\right]^2\left[ CrO _4^{-2}\right]$
$Ag _2 CrO _4 \rightleftharpoons\left[2 Ag ^{+}\right]\left[ CrO _4^{-2}\right]$
Hence, $K _{ sp }=\left[ Ag ^{+}\right]^2+\left[ CrO _4^{-2}\right]$
View full question & answer→MCQ 311 Mark
The solubility of an aqueous solution of $Mg ( OH )_2$ be $x$ then its $K _{ sp }$ is ............
- ✓
$4 x^3$
- B
$108 x^5$
- C
$27 x^4$
- D
$9 x$
AnswerCorrect option: A. $4 x^3$
(a) $4 x^3$
$
\begin{aligned}
Mg ( OH )_2 & \rightleftharpoons Mg ^{2+}+2 OH ^{-} \\
K _{\text {sp }} & =4 x^3 \text { }
\end{aligned}
$
View full question & answer→MCQ 321 Mark
The solubility product of a salt having a general formula $MX _2$ in water is $4 \times 10^{-2}$. The concentration of $M ^{2+}$ ions in the aqueous solution of the salt is ...............
- A
$2.0 \times 10^{-6} M$
- ✓
$1.0 \times 10^{-4} M$
- C
$1.6 \times 10^{-4} M$
- D
$4.0 \times 10^{-2} M$
AnswerCorrect option: B. $1.0 \times 10^{-4} M$
$(b)$ $1.0 \times 10^{-4} M$
Solution:
$MX _2 \rightleftharpoons M ^{2+}+2 X ^{-}$
$K _{ sp }=(2 s )^2( s )=4 s ^3 $
$s =2 \sqrt[3]{\frac{ K _{ sp }}{4}}=\sqrt[3]{\frac{4 \times 10^{-12}}{4}} \text { }$
$=1.0 \times 10^{-4} M$
View full question & answer→MCQ 331 Mark
On addition of ammonium chloride to a solution of ammonium hydroxide ……………
- ✓
dissociation of $NH _4 OH$ increases
- B
concentration of $OH ^{-}$increases
- C
concentration of $OH ^{-}$decreases
- D
concentration of $NH _4$ and $OH ^{-}$increases
AnswerCorrect option: A. dissociation of $NH _4 OH$ increases
dissociation of $NH _4 OH$ increases
View full question & answer→MCQ 341 Mark
A chemist dissolves an excess of $BaSO _4$ in pure water at $25^{\circ} C$ if its $K _{ sp }=1 \times 10^{-10}$ What is the concentration of Barium in the water?
- A
$10^{-14} M$
- B
$10^{-5} M$
- C
$10^{-15} M$
- ✓
$10^{-6} M$
AnswerCorrect option: D. $10^{-6} M$
$10^{-6} M$
View full question & answer→MCQ 351 Mark
Which of the following is not a Lewis acid?
- A
$CO$
- B
$SiCl _4$
- ✓
$SO _3$
- D
$Zn ^{2+}$
AnswerCorrect option: C. $SO _3$
$SO _3$
CO does not contain vacant d-orbital.
View full question & answer→MCQ 361 Mark
Three reactions involving $H _2 PO _4$ are given below.
(i) $H _3 PO _4+ H _2 O \rightarrow H _3 O ^{+}+ H _2 PO _4^{-}$
(ii) $H _2 PO _4+ H _2 O \rightarrow HPO _4{ }^2+ H _3 O ^{+}$
(iii) $H _2 PO _4+ OH ^{-} \rightarrow H _3 PO _4+ O _2^{-}$
In which of the above does $H _2 PO _4^{-}$act as an acid.
View full question & answer→MCQ 371 Mark
An acid $HA$ ionises as $HA \rightleftharpoons H ^{+}+ A ^{-}$The $pH$ of $1.0 M$ solution is 5 . its dissociation constant would be ..............
- A
$1 \times 10^{-5}$
- ✓
$1 \times 10^{-10}$
- C
- D
$5 \times 10^8$
AnswerCorrect option: B. $1 \times 10^{-10}$
$1 \times 10^{-10}$
View full question & answer→MCQ 381 Mark
At infinite dilution, the percentage ionisation for both strong and weak electrolyte is ………….
Answer(d) 100%
Hint:
According to Ostwald’s dilution law, degree of ionisation is directly proportional to the dilution.
View full question & answer→MCQ 391 Mark
Which of the following is non – electrolyte?
AnswerCorrect option: D. $CH _3 COOH$
$CH _3 COOH$
View full question & answer→MCQ 401 Mark
Which one of the following is called amphoteric solvent?
Answer(d) Water
Hint:
$H _2 O \rightarrow H ^{+}+ OH ^{-}$. Acid due to donation of proton.
Acid
$H _2 O + H ^{+} \rightarrow H _3 O ^{+}$. Base due to accepting a proton. So water is an amphoteric solvent.
Base
View full question & answer→MCQ 411 Mark
Among the following, the weakest Lewis base is …………
- A
$H ^{-}$
- B
$OH ^{-}$
- ✓
$Cl ^{-}$
- D
$HClO _3{ }^{-}$
AnswerCorrect option: C. $Cl ^{-}$
$Cl ^{-}$ is a conjugate base of strong acid $HCl$.
View full question & answer→MCQ 421 Mark
Which of the following is the weakest acid?
AnswerHF
Hint:
HF does not give proton easily.
View full question & answer→MCQ 431 Mark
Which of the following is the strongest Lewis acid?
- A
$BI _3$
- ✓
$BBr _3$
- C
$BCl _3$
- D
$BF _3$
AnswerCorrect option: B. $BBr _3$
Larger the size of the halogen atom less is the back donation of electrons into empty $2 p$ orbital of B.
View full question & answer→MCQ 441 Mark
Which one of the following substance has the highest proton affinity?
- A
$H _2 O$
- B
$H _2 S$
- ✓
$NH _3$
- D
$PH _3$
AnswerCorrect option: C. $NH _3$
$NH _3$
View full question & answer→MCQ 451 Mark
Which of the following is the strongest conjugate base?
- A
$Cl ^{-}$
- ✓
$CH _3 COO$
- C
$SO _4{ }^{2-}$
- D
$NO _2^{-}$
AnswerCorrect option: B. $CH _3 COO$
$CH _3 COO$ is a conjugate base of a weak acid.
$
CH _3 COOH \rightleftharpoons CH _3 COO + H ^{+}
$
View full question & answer→MCQ 461 Mark
Review the equilibrium and choose the correct statement.
$
HClO _4+ H _2 O \rightleftharpoons H _3 O ^{+}+ ClO _4^{+}
$
- A
$HClO _4$ is the conjugate acid of $H _2 O$
- B
$H _3 O$ is the conjugate base of $H _2 O$
- C
$H _2 O$ is the conjugate acid of $H _3 O$
- ✓
$ClO _4$ is the conjugate base of $HClO _4$
AnswerCorrect option: D. $ClO _4$ is the conjugate base of $HClO _4$
$ClO _4$ is the conjugate base of $HClO _4$
View full question & answer→MCQ 471 Mark
The unit of ionic product of water K is ……………
- A
$mol ^{-1} L ^{-1}$
- B
$mol ^{-2} L ^{-2}$
- C
$mol ^{-2} L$ ^{-1}
- ✓
$mol ^2 L ^{-2}$
AnswerCorrect option: D. $mol ^2 L ^{-2}$
$mol ^2 L ^{-2}$
View full question & answer→MCQ 481 Mark
The pH of millimolar HCl is ………….
Answer$(b) \ 3$
$pH =-\log \left[ H ^{+}\right]$
${\left[ H ^{+}\right]=10^{-3}}$
$pH =\log 1-\log \left[ H ^{+}\right]$
$=\log 1-\log 10^{-3}$
$=3$
View full question & answer→MCQ 491 Mark
As the temperature increases, the pH of KOH solution ……………..
- ✓
- B
- C
- D
depends upon concentration of KOH solution
View full question & answer→MCQ 501 Mark
Pure water is kept in a vessel and it remains exposed to atmospheric $CO _2$ which is absorbed, then its pH will be
- A
- B
- C
- ✓
depends on ionic production of water
AnswerCorrect option: D. depends on ionic production of water
depends on ionic production of water
Hint:
$CO _2$ is acidic oxide which on dissolution in water develops acidic nature.
View full question & answer→