Name: Condition for Precipitation
Uploaded: 2019-07-19
Duration: 6 min 35 s
Description: The video is based on the condition for precipitation of a salt dissolved in a solvent, where the ionic product should exceed the solubility product of the s...

Question

Calculate the hydrolysis constant of ${NH}_{4} Cl$ : determine the degree of hydrolysis of this salt in $0.01 M$ solution and the $pH$ of the solution.

$K_{b} ({NH}_{4} OH) = 1.8 \times 10^{- 5}$

Accepted Answer

${NH}_{4} Cl$ is a salt of weak base and strong acid.

Equation of hydrolysis is given as.

${NH}_{4}^{+} + H_{2} O ⇌ {NH}_{4} OH + H^{+} Initially c 0 0 Equillibrium c - ch ch ch K_{h} = \frac{[{NH}_{4} OH] [H^{+}]}{[{NH}_{4}^{+}]} . . . Equation (1) K_{h} = \frac{ch \times ch}{c (1 - h)} Since, h < < 1 . Then 1 - h \approx 1 K_{h} = {ch}^{2} . . . Equation (2)$

$Dissociation equation of weak base$
${NH}_{4} OH ⇌ {NH}_{4}^{+} + {OH}^{-} K_{b} = \frac{[{NH}_{4}^{+}] [{OH}^{-}]}{[{NH}_{4} OH]} . . . Equation (3) From equation 1 and 3 K_{h} = \frac{K_{w}}{K_{b}} K_{h} = \frac{1 \times 10^{- 14}}{1.8 \times 10^{- 5}} K_{h} = 5.5 \times 10^{- 10} Now put the value of K_{h} in equation 2 we get h = \sqrt{\frac{K_{h}}{c}} h = \sqrt{\frac{5.5 \times 10^{- 10}}{0.01}} h = 2.34 \times 10^{- 4}$

$Calculation of$ $p^{H}$

$[H^{+}] = ch [H^{+}] = 0.01 \times 2.34 \times 10^{- 4} [H^{+}] = 2.34 \times 10^{- 6} pH = - \log [H^{+}] pH = - \log [2.34 \times 10^{- 6}] pH = 5.64 Hence, pH = 5.64 .$

$Hydrolysis constant = 5.5 \times 10^{- 10}, Degree of hydrolysis = 2.4 \times 10^{- 4}, pH = 5.64$ .

Calculate the hydrolysis constant of ${NH}_{4} Cl$ : determine the degree of hydrolysis of this salt in $0.01 M$ solution and the $pH$ of the solution.

$K_{b} ({NH}_{4} OH) = 1.8 \times 10^{- 5}$

Important Questions on Ionic Equilibrium

Learn and Prepare for any exam you want !

Calculate the hydrolysis constant of NH4Cl: determine the degree of hydrolysis of this salt in 0.01 M solution and the pH of the solution. KbNH4OH=1.8×10-5

Important Questions on Ionic Equilibrium

Learn and Prepare for any exam you want !

Calculate the hydrolysis constant of ${NH}_{4} Cl$ : determine the degree of hydrolysis of this salt in $0.01 M$ solution and the $pH$ of the solution.

$K_{b} ({NH}_{4} OH) = 1.8 \times 10^{- 5}$