Chemistry - Chemical Equilibrium Question with Solution | TestHub

ChemistryChemical EquilibriumProblems involving more than one equilibriumMedium2 minQB

ChemistryMediumnumerical

Passage / Comprehension

HA is a weak acid, and BOH is a weak base. The ionization constant ( $K_{a}$ ) of HA is $3 \times 1 0^{- 7}$ , and the ionization constant ( $K_{b}$ ) of BOH is $3 \times 1 0^{- 7}$ at $298 K$ .

Percentage of hydrolysis of 0.1 M BA salt at 298 K is

Answer:

25.00

Solution:

$\frac{h}{1 - h} = K_{h}$

$\frac{h}{1 - h} = \frac{K _{w}}{K _{a} \cdot K _{b}}$

$\frac{h}{1 - h} = \frac{1 0 ^{- 14}}{3 \times 1 0 ^{- 7} \times 3 \times 1 0 ^{- 7}} \Rightarrow \frac{h}{1 - h} = \frac{1}{3}$

$3 h = 1 - h \Rightarrow 4 h = 1 \Rightarrow h = 0.25$

$P erce n t a g eo f h y d ro l ys i s = 25%$

Stream:JEESubject:ChemistryTopic:Chemical EquilibriumSubtopic:Problems involving more than one equilibrium

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...