Question

What amount of sodium propanoate should be added to one litre of an aqueous solution containing 0.02 mole of propanoic acid $(Ka=1.34\times {10}^{-5}at{25}^{0}C)$ to obtain a buffer solution of pH 4.75 :

• A. $4.52\times {10}^{-2}M$
• B. $3.52\times {10}^{-2}M$
• C. $2.52\times {10}^{-2}M$
• D. $1.52\times {10}^{-2}M$

Answer 1

The correct option is A $4.52\times {10}^{-2}M$

$pH=p{K}_a+log\frac{\left[salt\right]}{\left[acid\right]}$

Now pH of the solution is 4.75

$p{K}_a$ = $-log{K}_a$ 1.34 $\times$ ${10}^{-}$${}^5$ =4.870

[Acid] = 0.02mol/L

So,

4.75 = 4.870 + log $\frac{\left[salt\right]}{\left(0.02\right)}$

= log $\frac{\left[salt\right]}{\left(0.02\right)}$ = -0.12

= 0.8869

=$\left[salt\right]=0.8869\times 0.02$

= 0.017 mol/L

Now,

$K_a$ = $\frac{\left[H^{+}\right]\left[C_2{H}_{5}CO{O}^{-}\right]}{C_2{H}_{5}COOH}$

= $\frac{1.34\times {10}^5\times 0.007}{0.03}$

= $5.77\times {10}^{-}$${}^5$

= pH = -log$\left[H^{+}\right]$

= $4.52\times {10}^{-}$${}^2$ M