A buffer solution with pH=9 is to be prepared by mixing N H_4 Cl … - Vidyadip

MCQ

A buffer solution with $pH$$=9$ is to be prepared by mixing $N{H_4}Cl$ and $N{H_4}OH$. Calculate the number of moles of $N{H_4}Cl$ that should be added to one litre of $1.0\,M\;N{H_4}OH.\,$ $[{K_b} = 1.8 \times {10^{ - 5}}]$

A
$3.4$
B
$2.6$
C
$1.5$
✓
$1.8$

✓

Answer

Correct option: D.

$1.8$

(d) $pH = - {\rm{log}}\,\,{K_b} + {\rm{log}}\frac{{{\rm{[Salt]}}}}{{{\rm{[Acid]}}}}$

$pH = - {\rm{log}}\,[1.8 \times {10^{ - 5}}] + {\rm{log}}\frac{{{\rm{[Salt]}}}}{{{\rm{1}}{\rm{.0}}}}$

$9 = 4.7 + \log \frac{{{\rm{[Salt]}}}}{{1.0}}$ ; $\log \frac{{{\rm{[Salt]}}}}{{1.0}} = 4.7 - 9 = - 4.3$

$\frac{{{\rm{[Salt]}}}}{{1.0}} = {\rm{Antilog}}\frac{1}{{4.3}};\,\,[S{\rm{alt]}} = 1.8$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The molarity of $1 \mathrm{~L}$ orthophosphoric acid $\left(\mathrm{H}_3 \mathrm{PO}_4\right)$ having $70 \%$ purity by weight (specific gravity $\left.1.54 \mathrm{~g} \mathrm{~cm}^{-3}\right)$ is_____$\quad \mathrm{M}$.

(Molar mass of $\mathrm{H}_3 \mathrm{PO}_4=98 \mathrm{~g} \mathrm{~mol}^{-1}$ )

$FeCl_3$ on reaction with $K_4[Fe(CN)_6]$ in aqueous solution gives blue colour. These are separated by a semipermeable membrane $AB$ as shown. Due to osmosis there is

Monochlorination of toluene in presence of sunlight followed by hydrolysis gives

Assertion : When a metal is treated with conc. $HNO_3$ it generally yields a nitrate, $NO_2$
and $H_2O$.

Reason : Conc. $HNO_3$ reacts with metal and first produces a metal nitrate and nascent hydrogen. The nascent hydrogen then further reduces $HNO_3$ to $NO_2$.

The molar ionic conductances of the octahedral complexes.

$(1)$ $PtCl_4·5NH_3$ $(2)$ $PtCl_4·4NH_3$

$(3)$ $PtCl4·3NH_3$ $(4)$ $PtCl4·2NH_3$

$1-$ Alkyne and $2-$ alkyne can be distinguished by

Inorganic graphite is

The electrochemical equivalent $Z$ of any element can be obtained by multiplying the electrochemical equivalent of hydrogen with

Copper sulphate solution reacts with $KCN$ to give