Question

$\frac{N}{10}$ acetic acid was titrated with $\frac{N}{10}$ NaOH.When $25\%,50\%$ and $75$$\%$ of titration is over then the pH of the solution will be $:[K_a={10}^{-5}]$

• A. $5+\log 1/3,5,5+\log 3$
• B. $5+\log 3,4,5+\log 1/3$
• C. $5-\log 1/3,5,5-\log 3$
• D. $5-\log 1/3,4,5+\log 1/3$

Answer 1

The correct option is A $5+\log 1/3,5,5+\log 3$

t CH_3COOH NaOH CH_3COO^-Na^+

0 0.1 0.1

25% 0.1-0.025 0.1-0.025 0.025

50% 0.1-0.050 0.1-0.050 0.050

75% 0.1-0.075 0.1-0.075 0.075

$pH=-\log K_a+\log \frac{\left[salt\right]}{\left[acid\right]}$

t= 25%

$pH=5+\log \frac{0.025}{0.075}$

$pH=5+\log \frac{1}{3}$

t= 50%

$pH=5+\log \frac{0.050}{0.050}$

$pH=5+\log 1=5$

t= 75%

$pH=5+\log \frac{0.075}{0.025}$

$pH=5+\log 3$