Question

What will be the pH at the equivalence point during the titration of a $100$mL $0.2$ M solution of $C{H}_{2}COONa$ with $0.2$M solution of HCl ? $K_a=2$ $X$ ${10}^{-5}$

• A. $3-log\sqrt{2}$
• B. $3+log\sqrt{2}$
• C. $3-log2$
• D. $3+log2$

Answer 1

The correct option is A $3-log\sqrt{2}$

Volume of the solution $=100mL$ concentration $=0.2M$

$\begin{array}{c}{K}_a=2\times {10}^{-5}\\ {CH}_{3}COONa+HCl⟶{CH}_{3}COOH+NaCl\end{array}$

No. of millimoles of ${CH}_{3}COONa=C\times V$

$=0.2\times 100$

$=20$moles

No. of millimoles of $HCl=20m$ moles

$\begin{array}{cc}c=\frac{m}{v},\left[H^{+}\right]=\frac{20}{200}& \\ pH=-\frac{1}{2}(\log ka+\log c)& \\ =-\frac{1}{2}(\log (2\times {10}^{-5})+\log (0.1\left)\right)& \\ =& \frac{1}{2}(6-\log 2)\\ =& 3-\log \sqrt{2}\\ \text{(A) is correct option.}& \end{array}$