Question

100 mL of a 0.1 M $C{H}_{3}COOH$ is titrated with 0.1 M $NaOH$ solution. The $pH$ of the solution in the titration flask at the titre value of 50 is:

$\left[p{K}_a\right(C{H}_{3}COOH)=4.74]$

• A. 2.37
• B. 4.74
• C. 1.34
• D. 5.74

Answer 1

The correct option is B 4.74

$100$ ml of a $0.1$ M ${CH}_{3}COOH\equiv$ $10$ millimoles of ${CH}_{3}COOH$

now $50$ ml of $0.1$M $NaOH\equiv$ $5.0$ millimoles of $NaOH$.

$\therefore$ $10-5.0=5.0$ millimoles of ${CH}_{3}COONa$ will be produced.

$\therefore$ $pH=pKa+log\frac{\left[{CH}_{3}COONa\right]}{\left[{CH}_{3}COOH\right]}=4.74+log\left(\frac{5}{10-5}\right)=4.74$