You visited us 1 times! Enjoying our articles? Unlock Full Access!

Salt of Weak Acid and Weak Base

One litre of ...

Question

One litre of a buffer solution has $0.13833$ mole of acetic acid and $0.1951$ mole of sodium acetate. Calculate $p H$ (nearest integer) of solution. Neglect volume change.
$K_{a} = 1.8 \times 10^{- 5}$ . (log1.8 = 0.255)

Open in App

Solution

First, calculate the pKa value.

$p K a = - l o g K a = - l o g 1.8 \times 10^{- 5} = 4.75$

Now calculate the pH of the solution:

$p H = p K a + l o g \frac{s o d i u m a c e t a t e}{a c e t i c a c i d} = 4.75 + l o g (\frac{0.1951}{0.13833})$

$= 4.75 + 0.1440 = 4.8940 \approx 5$

Suggest Corrections

Similar questions

0.063

4.5

K_{a}

{C H}_{3} C O O H = 1.8 \times 10^{- 5}

K_{a}

{CH}_{3} COOH = 1.8 \times 10^{- 5}

\frac{N}{10}

K_{a} = 1.8 \times 10^{- 5}

H C l

K_{a} (C H_{3} C O O H) = 1.8 \times 10^{- 5}

l o g (1.8) = 0.255, l o g (0.66) = - 0.181

25^{o} C

p H = 5

K_{a}

1.8 \times 10^{- 5}

{log}_{10} 1.8 = 0.25 and 10^{(0.25)} = 1.8

Join BYJU'S Learning Program