Question

How many grams of ammonium chloride should be dissolved in 500 ml of water to have a solution of ${\text{pH}}$ 4.5?
[${{\text{K}}_{\text{b}}}$ for ammonium hydroxide is $1.8 \times {10^{ - 5}}$]
A) 25.22
B) 48.15
C) 53.5
D) 90

Answer 1

Write the dissociation reaction for ammonium chloride. Calculate the ${\text{p}}{{\text{K}}_{\text{b}}}$ from given ${{\text{K}}_{\text{b}}}$ and then calculate ${\text{p}}{{\text{K}}_a}$ from it. Use the Henderson Hassel balch to calculate the concentration of ammonium ion. Using the concentration of ammonium ion and volume of water calculate the moles of ammonium chloride. Finally, calculate the mass of ammonium chloride using moles and molar mass of it.

Formulas Used :
Henderson Hasselbalch equation
${\text{pH = p}}{{\text{K}}_{\text{a}}}{\text{ + log}}\dfrac{{{\text{[Base]}}}}{{{\text{[Acid]}}}}$
${\text{pH = }} - \log [{H^ + }]$
${\text{p}}{{\text{K}}_a} + {\text{p}}{{\text{K}}_{\text{b}}} = 14$

${\text{moles = molarity }} \times {\text{ vol L}}$

Complete step-by-step answer:
The dissociation reaction for ammonium chloride is as follows:
${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl(aq)}} \to {\text{N}}{{\text{H}}_{\text{3}}}{\text{(aq) + }}{{\text{H}}^{\text{ + }}}{\text{(aq)}}$
In this reaction ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$ is acidic species and ammonia is basic species.
We have given ${\text{pH}}$ a solution. Using ${\text{pH}}$ we can calculate the concentration of hydrogen ion as follows:
${\text{pH = }} - \log [{H^ + }]$
Now, substitute 4.5 for ${\text{pH}}$ and calculate the concentration of ${H^ + }$ ions.
$4.5 = - \log [{H^ + }]$
$[{H^ + }] = {\text{ antilog ( - 4}}{\text{.5) = 3}}{\text{.16}} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}{\text{M}}$
From the reaction, we can say that at equilibrium$[{H^ + }] = [{\text{N}}{{\text{H}}_{\text{3}}}{\text{] = 3}}{\text{.16}} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}{\text{M}}$.
Now use the Henderson Hasselbalch equation and calculate the concentration of ammonium ions. ${\text{pH = p}}{{\text{K}}_{\text{a}}}{\text{ + log}}\dfrac{{{\text{[Base]}}}}{{{\text{[Acid]}}}}$
We can rewrite it as
${\text{pH = p}}{{\text{K}}_{\text{a}}}{\text{ + log}}\dfrac{{{\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}}}{{{\text{[N}}{{\text{H}}_{\text{4}}}^ + {\text{]}}}}$
Here we need to calculate ${\text{p}}{{\text{K}}_a}$ from the given ${{\text{K}}_{\text{b}}}$ value of ammonium hydroxide.
${\text{p}}{{\text{K}}_{\text{b}}} = - \log ({{\text{K}}_{\text{b}}}) = - \log (1.8 \times {10^{ - 5}}) = 4.74$
Using the relation between ${\text{p}}{{\text{K}}_{\text{b}}}$ and ${\text{p}}{{\text{K}}_a}$ we can calculate the value of ${\text{p}}{{\text{K}}_a}$ as follows :
${\text{p}}{{\text{K}}_a} + {\text{p}}{{\text{K}}_{\text{b}}} = 14$
So, ${\text{p}}{{\text{K}}_a} = 14 - {\text{p}}{{\text{K}}_{\text{b}}} = 14 - 4.74 = 9.26$
Now, substitute 4.5 for${\text{pH}}$, 9.26 for ${\text{p}}{{\text{K}}_a}$ , ${\text{3}}{\text{.16}} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}{\text{M}}$ for $[{\text{N}}{{\text{H}}_{\text{3}}}{\text{]}}$ and calculate the equilibrium concentration of ${\text{[N}}{{\text{H}}_{\text{4}}}^ + {\text{]}}$ using the Henderson Hasselbalch equation.
$\Rightarrow {\text{4}}{\text{.5 = 9}}{\text{.26 + log}}\dfrac{{{\text{3}}{\text{.16}} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}{\text{M}}}}{{{\text{[N}}{{\text{H}}_{\text{4}}}^ + {\text{]}}}}$
$\Rightarrow {\text{ - 4}}{\text{.76 = log}}\dfrac{{{\text{3}}{\text{.16}} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}{\text{M}}}}{{{\text{[N}}{{\text{H}}_{\text{4}}}^ + {\text{]}}}}$
$\Rightarrow \dfrac{{{\text{3}}{\text{.16}} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}{\text{M}}}}{{{\text{[N}}{{\text{H}}_{\text{4}}}^ + {\text{]}}}} = {\text{antilog( - 4}}{\text{.76}})$
$\Rightarrow \dfrac{{{\text{3}}{\text{.16}} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}{\text{M}}}}{{{\text{[N}}{{\text{H}}_{\text{4}}}^ + {\text{]}}}} = 1.74 \times {10^{ - 5}}$
$\Rightarrow {\text{[N}}{{\text{H}}_{\text{4}}}^ + {\text{]}} = 1.82{\text{M}}$

We will consider this as the initial concentration of ammonium chloride as it dissociates very less.
Now we will calculate the moles of ammonium chloride using the volume of water and calculated concentration.
$\Rightarrow {\text{moles = molarity }} \times {\text{ vol L}}$
Volume off water = 500 ml =0.5L
$\Rightarrow {\text{moles N}}{{\text{H}}_{\text{4}}}{\text{Cl = 1}}{\text{.82M}} \times {\text{ 0}}{\text{.5L = 0}}{\text{.91 mole}}$
Now, we will convert these moles of ammonium chloride to mass using its molar mass as follows:
The molar mass of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$=${\text{53}}{\text{.5g/mol}}$
$\Rightarrow {\text{mass N}}{{\text{H}}_{\text{4}}}{\text{Cl = 0}}{\text{.91 mole}} \times {\text{53}}{\text{.5g/mol}} = {\text{48 g}}$
Hence, we can say that 48.15 g ammonium chloride should be dissolved in 500 ml of water to have a solution of ${\text{pH}}$ 4.5.

Thus, the correct answer is option (B) 48.15.

Note: Acids are proton donor species. The ability of acid to donate the proton determines the strength of the acid. Strong acid dissociates completely in an aqueous solution. Weak acid dissociates partially in an aqueous solution. As weak acid dissociates very less we consider initial concentration and equilibrium concentration of weak acid same.