The ionization constant of nitrous acid is 4-5- 10 -4- Calculate the pH of 0-04 M sodium nitrite solution and also its degree of hydrolysis-

pK_a = log K_a = log (4.5 × 10^ 4) =3.35 pH = 0.5pK_w + 0.5 pK_a + log c = 0.5 × 14 + 0.5 × 3.35 + log 0.04 = 7.285 K_h = √(K_w/K_ac) = √(10^ 14 ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Salt of Strong Acid and Weak Base

The ionizatio...

Question

The ionization constant of nitrous acid is $4.5 \times 10^{- 4}$ . Calculate the $p H$ of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Open in App

Solution

$p K_{a} = - l o g K_{a} = - l o g (4.5 \times 10^{- 4}) = 3.35$
$p H = 0.5 p K_{w} + 0.5 p K_{a} + l o g c = 0.5 \times 14 + 0.5 \times 3.35 + l o g 0.04 = 7.285$
$K_{h} = \sqrt{\frac{K_{w}}{K_{a} c}} = \sqrt{\frac{10^{- 14}}{4.5 \times 10^{- 4} \times 0.04}} = 2.35 \times 10^{- 5}$

Suggest Corrections

Similar questions

The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

The ionization constant of propanoic acid is $1.32 \times 10^{- 5}$ . Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

Q. Calculate the degree of hydrolysis and pH of

0.1 M

sodium acetate solution. Hydrolysis constant of sodium acetate is

5.6 \times 10^{- 10}

Join BYJU'S Learning Program

The ionization constant of nitrous acid is 4.5×10−4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

pKa=−logKa=−log(4.5×10−4)=3.35pH=0.5pKw+0.5pKa+logc=0.5×14+0.5×3.35+log0.04=7.285Kh=√KwKac=√10−144.5×10−4×0.04=2.35×10−5

The ionization constant of nitrous acid is $4.5 \times 10^{- 4}$ . Calculate the $p H$ of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

$p K_{a} = - l o g K_{a} = - l o g (4.5 \times 10^{- 4}) = 3.35$
$p H = 0.5 p K_{w} + 0.5 p K_{a} + l o g c = 0.5 \times 14 + 0.5 \times 3.35 + l o g 0.04 = 7.285$
$K_{h} = \sqrt{\frac{K_{w}}{K_{a} c}} = \sqrt{\frac{10^{- 14}}{4.5 \times 10^{- 4} \times 0.04}} = 2.35 \times 10^{- 5}$