Question

Calculate the pH at the equivalence point when a solution of $0.1$ M acetic acid is titrated with a solution of $0.1$ M sodium hydroxide.${K_a}$ for acetic acid= $1.9 \times {10^{ - 5}}$

Answer 1

Hint: At first we will write what is given in the question. Then we will find the concentration of sodium acetate used at equal volumes. . Then we will see reaction at equilibrium. Then we will calculate the hydrolysis constant of the equation. From there we will calculate the amount dissociated. We will find the concentration of hydrogen and hydride ion. We can calculate the pH from hydrogen ion concentration.

Complete step-by-step solution:Step1. Molarity of the acetic acid is $0.1$ M
Molarity of the sodium hydroxide: $0.1$ M
${K_w}$: $1 \times {10^{ - 14}}$ ;w is water
${K_a}$ for acetic acid= $1.9 \times {10^{ - 5}}$
WE need to calculate the pH.
Step2. The concentration of sodium acetate which is formed when acetic acid and sodium hydroxide are mixed will be $\dfrac{{0.1}}{2} = 0.5M$ at equal volume of both acid and base is considered.
Step3. The equilibrium reaction is given below:
\[C{H_3}CO{O^ - } + {H_2}O \rightleftharpoons C{H_3}COOH + O{H^ - }\]
. So the concentration of $C{H_3}CO{O^ - }$ is $C(1 - x)$ and of $C{H_3}COOH$ is Cx and of $O{H^ - }$will be also Cx where the x is the degree of hydrolysis.
Step4. The hydrolysis constant will be ${K_h} = \dfrac{{C{x^2}}}{{1 - x}}$
So now we know that
${K_h} = \dfrac{{{K_w}}}{{{K_a}}} = \dfrac{{1.0 \times {{10}^{ - 14}}}}{{1.9 \times {{10}^{ - 5}}}} = 5.26 \times {10^{ - 10}}$
Now
$
{K_h} = C{x^2} \\
\Rightarrow 5.26 \times {10^{ - 10}} = 0.05 \times {x^2} \\
\therefore x = 1.025 \times {10^{ - 4}} \\
\ $
Here we got the x which is the degree of hydrolysis.
Step5. We will calculate the concentration of hydroxide ions.
$O{H^ - } = Cx = 1.025 \times {10^{ - 4}} \times 0.05 = 5.125 \times {10^{ - 6}}M$
Now in water there are two types of ions: one is hydrogen and other is hydroxide ion so we can find the hydrogen ion concentration by using the ${K_w}$ divided by the hydroxide concentration.
${H^ + } = \dfrac{{1 \times {{10}^{ - 14}}}}{{5.125 \times {{10}^{ - 6}}}} = 1.95 \times {10^{ - 9}}M$
Step6. Now the formula of pHis the negative of log of ${H^ + }$
$pH = - \log [{H^ + }] = 8.71$

Hence the pH is the $8.71$.

Note:pH of the solution is used to determine the nature of the solution. If the pH is greater than seven that is the basic solution and if it is less than seven then it is an acidic solution.