Question

Calculate the $pH$ of buffer solution with $0.1{\text{ M}}$ each of $C{H_3}COOH$ and $C{H_3}COONa$ . What will be changed in $pH$ in addition to the following?
$(i){\text{ 0}}{\text{.01 M HCl}}$
$(ii){\text{ 0}}{\text{.01 M NaOH}}$
The total volume of solution is ${\text{1 L}}$ . $\left( {{{\text{K}}_a}{\text{ = 1}}{\text{.8 }} \times {\text{ 1}}{{\text{0}}^{ - 5}}} \right)$

Answer 1

Buffer solution is the solution which has no effect on the $pH$ of the solution on small addition of acid or base to it. We will find the pH of the buffer by using ${{\text{K}}_a}$ and the given concentration of acetic acid and sodium acetate. Then we will increase the concentration of both in addition to ${\text{HCl}}$ and $NaOH$ .
$pH{\text{ }} = {\text{ }}p{K_a}{\text{ }} + {\text{ }}\log \left[ {\dfrac{{conjugate{\text{ }}base}}{{weak{\text{ }}acid}}} \right]$
Where ${K_a}$ is dissociation constant.

Complete answer:
A buffer solution containing $0.1{\text{ M}}$ of $C{H_3}COOH$ and $0.1{\text{ M}}$ of $C{H_3}COONa$ with ${\text{p}}{{\text{K}}_a}{\text{ = 1}}{\text{.8 }} \times {\text{ 1}}{{\text{0}}^{ - 5}}$ is given. Its $pH$ can be calculated as:
$pH{\text{ }} = {\text{ }}p{K_a}{\text{ }} + {\text{ }}\log \left[ {\dfrac{{conjugate{\text{ }}base}}{{weak{\text{ }}acid}}} \right]$
$\Rightarrow pH{\text{ }} = {\text{ }}p{K_a}{\text{ }} + {\text{ }}\log \left[ {\dfrac{{C{H_3}COONa}}{{C{H_3}COOH}}} \right]$
We know that $p{K_a}{\text{ }} = {\text{ }} - \log \left[ {{K_a}} \right]$ , therefore it can be deduced as;
$\Rightarrow pH{\text{ }} = {\text{ }} - \log \left[ {{K_a}} \right]{\text{ }} + {\text{ }}\log \left[ {\dfrac{{C{H_3}COONa}}{{C{H_3}COOH}}} \right]$
On substituting the values we get the result as,
$\Rightarrow pH{\text{ }} = {\text{ - log}}\left[ {{\text{1}}{\text{.8 }} \times {\text{ 1}}{{\text{0}}^{ - 5}}} \right]{\text{ }} + {\text{ }}\log \left[ {\dfrac{{0.1}}{{0.1}}} \right]$ ____________ $(1)$
$\Rightarrow pH{\text{ }} = {\text{ - log}}\left( {{\text{1}}{\text{.8 }} \times {\text{ 1}}{{\text{0}}^{ - 5}}} \right){\text{ }} + {\text{ }}\log \left( 1 \right)$
We know that, $\log (1) = 0$ , therefore,
$\Rightarrow pH{\text{ }} = {\text{ - log}}\left( {{\text{1}}{\text{.8 }} \times {\text{ 1}}{{\text{0}}^{ - 5}}{\text{ }}} \right)$
$\Rightarrow pH{\text{ }} = {\text{ 5 - log(1}}{\text{.8)}}$
$\Rightarrow pH{\text{ }} = {\text{ 4}}{\text{.74}}$
$(i){\text{ 0}}{\text{.01 M HCl}}$
Now when we add ${\text{ 0}}{\text{.01 M HCl}}$ then concentration acid will be $\left( {0.1 + 0.01} \right)M$ . Therefore the $pH$ will be calculated as:
$pH{\text{ }} = {\text{ }}p{K_a}{\text{ }} + {\text{ }}\log \left[ {\dfrac{{conjugate{\text{ }}base}}{{weak{\text{ }}acid}}} \right]$
$pH{\text{ }} = {\text{ }}p{K_a}{\text{ }} + {\text{ }}\log \left[ {\dfrac{{0.1}}{{0.01 + 0.1}}} \right]$
Since $0.01 \ll 0.1$ , we can write as:
$\Rightarrow 0.01{\text{ }} \ll {\text{ }}0.1{\text{ }} \approx {\text{ }}0.1$
Thus the equation reduced as:
$\Rightarrow pH{\text{ }} = {\text{ - log}}\left[ {{\text{1}}{\text{.8 }} \times {\text{ 1}}{{\text{0}}^{ - 5}}} \right]{\text{ }} + {\text{ }}\log \left[ {\dfrac{{0.1}}{{0.1}}} \right]$
Which is nothing but equation $(1)$ . Therefore it $pH$ will be equal to:
$\Rightarrow pH{\text{ }} = {\text{ 4}}{\text{.74}}$
$(ii){\text{ 0}}{\text{.01 M NaOH}}$
Now the concentration of the conjugate base will be $\left( {0.1 + 0.01} \right)M$ . Therefore it $pH$ will be calculated as:
$pH{\text{ }} = {\text{ }}p{K_a}{\text{ }} + {\text{ }}\log \left[ {\dfrac{{conjugate{\text{ }}base}}{{weak{\text{ }}acid}}} \right]$
$pH{\text{ }} = {\text{ }}p{K_a}{\text{ }} + {\text{ }}\log \left[ {\dfrac{{0.01 + 0.1}}{{0.1}}} \right]$
Since we know that $0.01 \ll 0.1$ , we can write as:
$\Rightarrow pH{\text{ }} = {\text{ - log}}\left[ {{\text{1}}{\text{.8 }} \times {\text{ 1}}{{\text{0}}^{ - 5}}} \right]{\text{ }} + {\text{ }}\log \left[ {\dfrac{{0.1}}{{0.1}}} \right]$
Which is again nothing but equation $(1)$ . Therefore it $pH$ will be equal to:
$\Rightarrow pH{\text{ }} = {\text{ 4}}{\text{.74}}$
Thus the buffer has no change in $pH$ in addition to a small amount of acid or base.

Note:
It must be noted while neglecting the molar concentration, it must be lesser or greater by factor ten. Then only it can be neglected. Also a buffer solution is a buffer in addition to base or acid to it. Thus it shows a very small change in $pH$ which can be neglected easily. This is the basic property of the buffer solution.