Question

What will be the $pH$ at the equivalence point during the titration of a $100 \,mL \,0.2 \,M$ solution of $CH _{3} COONa$ with $0.2 \,M$ solution of $HCl $ ? $K_{a}=2 \times 10^{-5}$

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

Answer 1

Answer: (A)

$CH _{3} COONa + HCl \longrightarrow CH _{3} COOH + NaCl$

at equivalence the $\left[ CH _{3} COOH \right]=\frac{20}{200} \Rightarrow 0.1$

$pH =\frac{1}{2} p K_{a}-\frac{1}{2} \log C$

$pH =\frac{1}{2}\left[5 \log 2-\log 10^{-1}\right]$

$pH =\frac{1}{2}[6 \log 2] \Rightarrow pH =3-\log \sqrt{2}$