Question

For ${\text{500 mL}}$ of ${\text{0}} \cdot {\text{2 M}}$ aqueous solution of acetic acid is mixed with ${\text{500 mL}}$ of ${\text{0}} \cdot {\text{2 M HCl}}$ at ${25^o}C$. If ${\text{6 g}}$ of ${\text{NaOH}}$ is added to the above solution, determine the final pH (assuming no change in volume on mixing. ${K_a}$ of acetic acid is $1 \cdot 75 \times {10^{ - 5}}mol{L^{ - 1}}$. (Write the value of nearest integer)

Answer 1

Hint:First calculate the number of moles present of ${\text{NaOH}}$. To know the value of pH one must always first try to find the value of $p{K_a}$. Once if you know the value of $p{K_a}$ you can directly substitute the values in the formulae to identify the pH.

Complete answer:
1) First of all we will calculate the number of moles of ${\text{NaOH}}$ as below,
Number of moles of ${\text{NaOH}}$ $ = \dfrac{{{\text{Given weight}}}}{{{\text{Molecular weight of NaOH}}}}$
Weight is given of ${\text{NaOH = 6 grams}}$
Molecular weight of ${\text{NaOH = 23 + 16 + 1 = 40 grams}}$
Now let us put these values in the above formula we get,
Number of moles of ${\text{NaOH}}$ $ = \dfrac{{{\text{Given weight}}}}{{{\text{Molecular weight of NaOH}}}}$
Number of moles of ${\text{NaOH}}$ $ = \dfrac{6}{{40}}$
Number of moles of ${\text{NaOH}}$ $ = 0 \cdot 15moles$
2) Now out of $0 \cdot 15moles$ of ${\text{NaOH}}$ ${\text{0}} \cdot {\text{1M}}$ will be neutralized with ${\text{HCl}}$ (which is a strong acid) and ${\text{0}} \cdot {\text{005M}}$ will be neutralized with acetic acid (which is a weak acid). Hence, we can say that,
$\left[ {C{H_3}COOH} \right] = 0 \cdot 05M$ and $\left[ {C{H_3}COONa} \right] = 0 \cdot 05M$
3) As mentioned before now we need to find out the value of pH and for that, we need the value of $p{K_a}$ as below,
$p{K_a} = - {\log _{10}}\left[ {{K_a}} \right]$
Now the value of ${K_a}$ is given as $1 \cdot 75 \times {10^{ - 5}}$ and lets put that value in the above equation,
$p{K_a} = - {\log _{10}}\left[ {1 \cdot 75 \times {{10}^{ - 5}}} \right]$
After calculating the above logarithmic value we get,
$p{K_a} = 4 \cdot 75$
4) As we know the $p{K_a}$ value we can calculate the pH by using the following formula,
$pH = p{K_a} + \log \left[ {C{H_3}COOH/C{H_3}COONa} \right]$
By putting the known values in the above equation we get,
$pH = 4 \cdot 75 + \log \left[ {0 \cdot 05/0 \cdot 05} \right]$
$pH = 4 \cdot 75 + \log \left[ 1 \right]$
As the value of $\log \left[ 1 \right] = 0$ we get,
$pH = 4 \cdot 75 + 0$
$pH = 4 \cdot 75$
Therefore, the pH value is ${\text{4}} \cdot {\text{75}}$ and the nearest integer value is ${\text{5}}$. Hence, the pH is ${\text{5}}$.

Note:
Always remember to find out the number of moles for the given compound when gram addition is done in a solution. Finding the $p{K_a}$ value before pH will make it easy to find out the value of pH.