Question

Calculate the pH of a buffer solution prepared by dissolving $\text{ 30 g }$ of $\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$ in $\text{ 500 mL }$ of an aqueous solution containing $\text{ 150 mL }$ of$\text{ 1M HCl }$. $\text{ }{{\text{K}}_{\text{a}}}\text{ }$ For: $\text{ HC}{{\text{O}}_{\text{3}}}\text{ = 5}\text{.63 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-11}}}\left[ \text{log}\left( \dfrac{\text{133}}{\text{150}} \right)\text{ = }-\text{0}\text{.05 } \right]\text{ }$
(A) $\text{ }8.25\text{ }$
(B) $\text{ }9.19\text{ }$
(C) $\text{ }10.43\text{ }$
(D) $\text{ }11.56\text{ }$

Answer 1

A buffer solution is one which resists changes in pH when small quantities of an acid or alkali added to it. The formula to find the pH of a buffer solution is as below,$\text{ pH of buffer = p}{{\text{K}}_{\text{a}}}\text{+log}\dfrac{\text{ }\\!\\![\\!\\!\text{ salt }\\!\\!]\\!\\!\text{ }}{\text{ }\\!\\![\\!\\!\text{ base }\\!\\!]\\!\\!\text{ }}\text{ }$
- Where the $\text{ p}{{\text{K}}_{\text{a}}}\text{ }$ is a negative logarithmic value of dissociation constant${{\text{K}}_{\text{a}}}$.

Complete Solution:
First of all, let’s see how sodium carbonate will react with $\text{ HCl }$ to adjust the concentration of hydrogen ions in the system.
- Hydrochloric acid reacts with sodium carbonate to neutralize the base as follows: $\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ + HCl }\to \text{ NaCl + }{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$

Now, we will find the concentration of both to find pH from the formula given in the hint part. So, we can say that,
$\text{ Moles of N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ = }\dfrac{\text{Weight}}{\text{Molecular weight}}=\dfrac{\text{m}}{\text{M}}\text{ = }\dfrac{\text{30}}{\text{105}}\text{ = 0}\text{.28 mol }$
Now, we are not given the weight of the solution directly but we can find the number of its moles from its molarity by the following equation.
$\text{ Molarity = }\dfrac{\text{Moles of solute}}{\text{Volume of solution(in L)}}\text{ }$

So, we can write that,
$\begin{aligned} & \text{ Moles of HCl = Molarity }\\!\\!~\\!\\!\text{ (M) }\times \text{Volume of the solution (V}) \\\ & \text{ }\therefore \text{Moles of HCl = 1 }\\!\\!~\\!\\!\text{ }\\!\\!\times\\!\\!\text{ }\left( \text{0}\text{.150 L} \right)\text{ = 0}\text{.15 mol} \\\ \end{aligned}$
So, Moles of $\text{ HCl }$is equal to the$\text{ 0}\text{.15 mol}$.
So, from this, we can say that $\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$ will react with all the available $\text{ HCl }$ in the solution. Therefore, 0.15 moles of $\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$ will be used.

Hence we can write:
The final concentration of $\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$ after reaction:
$\begin{aligned} & \text{ Moles of N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{= Initial moles }-\text{ number of moles reacted with HCl} \\\ & \therefore \text{Moles of N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ = }0.28\text{ }-\text{ }0.15\text{ }=\text{ }0.13\text{ mole} \\\ \end{aligned}$

The concentration of $\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$ can be calculated as follows:
$\text{ Concentration = }\dfrac{\text{Number of moles}}{\text{Total volume of the solution}}$
So, we can write that:
$\text{ Concentration of N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ = }\dfrac{\text{0}\text{.13}}{\left( \text{0}\text{.500 + 0}\text{.15} \right)}\text{M = 0}\text{.252M}$
And the salt ($\text{ NaCl }$ ) produced will also have the same number of moles of $\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$ used. So, we can say that,
$\text{ Concentration of salt = }\dfrac{\text{0}\text{.15}}{\left( \text{0}\text{.5 + 0}\text{.15} \right)}\text{ M = 0}\text{.230M}$

- Now, we can use the formula of the pH of the buffer given below:
$\text{ pH of buffer = p}{{\text{K}}_{\text{a}}}\text{+}\left( \dfrac{\text{log}\left[ \text{Salt} \right]}{\left[ \text{Base} \right]} \right)\text{ }$
$\text{ Now, p}{{\text{K}}_{a}}\text{ = }-\text{log }\left[ {{\text{K}}_{\text{a}}} \right]\text{ = }-\text{log }\left[ \text{5}\text{.63 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-11}}} \right]\text{ =10}\text{.32}$

So, we can write equation as:
$\text{ pH = 10}\text{.32 + log}\left[ \dfrac{\text{0}\text{.15}}{\text{0}\text{.13}} \right]\text{ = 10}\text{.32 + 0}\text{.11 = 10}\text{.43}$
Thus, the pH buffer solution prepared by dissolving$\text{ 30 g }$ of $\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$ in $\text{ 500 mL }$ of an aqueous solution containing $\text{ 150 mL }$ of $\text{ 1M HCl }$ is 10.43

So, the correct answer is “Option C”.

Note: According to the Henderson-Hasselbach equation, when the concentration of acid and its conjugate base is equal that is when the acid is $\text{ }50{\scriptstyle{}^{0}/{}_{0}}\text{ }$ dissociated in the solution, then the $\text{pH}$of the solution would be,
$\begin{aligned} & \text{ pH = p}{{\text{K}}_{\text{a}}}\text{+log}\dfrac{\text{ }\\!\\![\\!\\!\text{ salt }\\!\\!]\\!\\!\text{ }}{\text{ }\\!\\![\\!\\!\text{ base }\\!\\!]\\!\\!\text{ }}\text{ } \\\ & \text{pH = p}{{\text{K}}_{\text{a}}}\text{+log (1) }\because \text{ log 1 =0} \\\ & \therefore \text{ pH = p}{{\text{K}}_{\text{a}}} \\\ \end{aligned}$