The degree of hydrolysis h of 0.020 M NH4Cl (K_b = 1.8x 10^-5) is calculated from h = sqrtK_h/C. Report hx 10^4 to two decimals.

JEE Main levelNumerical answerDifficulty 2 of 3ia120

The degree of hydrolysis $h$ of $0.020 M$ $NH X_{4} Cl$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ) is calculated from $h = K_{h} / C$ . Report $h \times 1 0^{4}$ to two decimals.

Numerical answer type, tolerance 0.01.

K_{h} = K_{w} / K_{b} = 5.56 \times 1 0^{- 10}

h = 5.56 \times 1 0^{- 10} /0.020 = 1.667 e - 04

, so

h \times 1 0^{4} = 1.67