Question

How do you calculate the change in $ pH $ when $ 3.00mL $ of $ {\text{0}}{\text{.100}}M $ $ HCL(aq) $ is added to $ 100.0mL $ of a buffer solution that is $ 0.100M $ in $ N{H_3}(aq) $ and $ 0.100M $ in $ N{H_4}Cl(aq) $ ?

Answer 1

Hint: First, the initial $ pH $ of the buffer solution along with the number of moles of acid and base should be calculated. After that, the change in the number of moles will be calculated which happens due to the addition of $ HCl $ in the buffer solution. Then the new $ pH $ of the solution should be calculated. Now to calculate change in $ pH $ , just subtract the initial $ pH $ from the final $ pH $ of the buffer solution.

Complete step by step solution:
First we will calculate the initial $ pH $ of the solution using Henderson-Hasselbalch equation, which states that for a buffer solution that contains a weak base and its conjugate acid,
 $ pOH = p{K_b} + \log \dfrac{{{\text{[conjugate acid]}}}}{{[{\text{weak base}}]}} $
The initial concentration of the both weak base and its conjugate acid is $ 0.100M $ . On substituting the above values into the equation, we get,
 $ pOH = p{K_b} + \log \dfrac{{{\text{0}}{\text{.100 M}}}}{{0.100{\text{ M}}}} $
 $ pOH = p{K_b} $
We know that the values of $ p{K_b}{\text{ (of }}N{H_3}{\text{)}} = 4.75 $ , therefore,
 $ pOH = 4.75 $
Thus,
 $ pH = 14 - pOH $
 $ pH = 14 - 4.75 $
Which results in,
 $ pH = 9.25 $
Hence, the $ pH $ of the buffer solution is $ 9.25 $ .
In addition to $ HCl $ in the buffer solution, it reacts with the weak base to produce more conjugate acid.
 $ HC{l_{(aq)}} + N{H_3}_{(aq)} \to N{H_4}Cl{ _{(aq)}} $
The above reaction tells that $ HCl $ is completely consumed in the process.
First, let's calculate the original number of moles of acid and base.
Number of moles of $ N{H_4}Cl = $ Molarity $ \times $ volume $ = 0.100M \times 0.1L $
Number of moles of $ N{H_4}Cl = 0.01mol $
Similarly,
Number of moles of $ N{H_3} = $ Molarity $ \times $ volume $ = 0.100M \times 0.1L $
Number of moles of $ N{H_3} = 0.01mol $
And the number of moles of $ HCl $ that will be added in the solution,
Number of moles of $ HCl = $ Molarity $ \times $ volume $ = 0.100M \times 0.003L $
Number of moles of $ HCl = 0.0003mol $
The above calculation indicates that there is enough $ HCl $ present to consume $ 0.0003mol $ of the base to produce $ 0.0003mol $ of acid from the buffer solution.
Hence,
 $ 0.01{\text{ }}mol{\text{ }}N{H_3} - 0.0003{\text{ }}mol{\text{ = }}0.0097{\text{ }}mol{\text{ }}N{H_3} $
 $ 0.01{\text{ }}mol{\text{ }}N{H_4}Cl + 0.0003{\text{ }}mol{\text{ = }}0.0103{\text{ }}mol{\text{ }}N{H_4}Cl $
Now, let's calculate the new $ pH $ of the solution after addition of $ HCl $ using Henderson-Hasselbalch equation.
 $ pOH = p{K_b} + \log \dfrac{{[N{H_4}Cl]}}{{[N{H_3}]}} $
 $ pOH = 4.75 + \log \dfrac{{0.0103}}{{0.0097}} $
On further solving,
 $ pOH = 4.75 + 0.0261 $
 $ pOH = 4.78 $
And thus,
 $ pH = 14 - pOH $
 $ pH = 14 - 4.78 $
Which results in,
 $ pH = 9.22 $
Now, change in $ pH $ can be calculated as
 $ \Delta = p{H_{initial}} - p{H_{final}} $
 $ \Delta = 9.22 - 9.25 $
 $ \Delta = - 0.03 $
Hence, we can say that there is a decrease in $ pH $ by $ - 0.03 $ when $ HCl $ is added.

Note:
When calculating the initial and final $ pH $ of the solution, pay close attention to whether you are using $ p{K_a} $ or $ p{K_b} $ in Henderson-Hasselbalch equation, as both of them have different values and carelessly using one in place of other will result in wrong value of $ pH $ . Also, while calculating the concentration of acid and base after addition of $ HCl $ , remember to add and subtract the change in number of moles of the concentration carefully.