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

Hint: 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.