MCQ
What will be the resultant $pH$ when $100\,ml$ of an aqueous solution of $HCl\, (pH = 2)$ is mixed with $200\,ml$ of an aqueous solution of $NaOH$ $(pH = 12)$
- A$2.52$
- ✓$11.52$
- C$3.30$
- D$4$
$\therefore\left[\mathrm{H}^{+}\right]=10^{-2} \,\mathrm{M}$
$\mathrm{pH}$ of $\mathrm{NaOH}=12$
$\therefore\left[\mathrm{OH}^{-}\right]= 10^{-2} \,\mathrm{M}$
$\mathrm{HCl}+\mathrm{NaOH} \longrightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}$
Initial mili mol
$100\times10^{-2}$ $200\times10^{-2}$
$1$ $2$ $0$ $0$
Final mili mol $0$ $1$ $1$ $1$
$[\mathrm{OH}]$ from $\mathrm{NaOH}=\frac{1}{300}=3 \times 10^{-3}$
$\mathrm{pOH} =-\log \left[\mathrm{OH}^{-}\right]=-\log \left[3 \times 10^{-3}\right]$
$=-\log \left[3 \times 10^{-3}\right]$
$=-[0.4771-3]$
$=2.52$
$\mathrm{pH}=14 -2.52=11.48$
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[Given : mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$, Plank's constant $(\mathrm{h})=6.626 \times 10^{-34} \mathrm{JS}$ ]
(Value of $\pi=3.14$ )
$CH_3-CH_2-C \equiv CH + HCl \to B \xrightarrow{{HCl}} C$
|
LIST $I$ (Precipitating reagent and conditions) |
LIST $II$ (Cation) |
| $A$ $\mathrm{NH}_4 \mathrm{Cl}+\mathrm{NH}_4 \mathrm{OH}$ | $I$ $\mathrm{Mn}^{2+}$ |
| $B$ $\mathrm{NH}_4 \mathrm{OH}+\mathrm{Na}_2 \mathrm{CO}_3$ | $II$ $\mathrm{Pb}^{2+}$ |
| $C$ $\mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl}+\mathrm{H}_2 \mathrm{~S}$ gas | $III$ $\mathrm{Al}^{3+}$ |
| $D$ dilute $\mathrm{HCl}$ | $IV$ $\mathrm{Sr}^{2+}$ |
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