Question

A 0.500 m solution of a weak base has a ph of 9.75. what is the base hydrolysis constant, kb, for the weak base?

Answer 1

when we have PH = 9.75
So we can get POH = 14 - 9.75 = 4.25 
and when POH = - ㏒[OH-]
by substitution:
4.25 = -㏒[OH-]
∴[OH] = 5.6x10^-5
from this reaction equation:
BOH ↔ B+  + OH-
∴[OH-] = [B+]= 5.6x10^-5 M
and Equ [BOH] = 0.5 m - X
                          = 0.5 - (5.6x10^-5)
                          =  0.4999
∴ Kb = [OH-][B+]/[BOH]
         = (5.6x10^-5)^2 / 0.4999
         = 6.27x10^-9

Answer 2

The hydrolysis constant for the hydrolysis reaction of the weak base is the

equilibrium constant for the reaction.

The hydrolysis constant is approximately 1.581 × 10⁻⁶

Reasons:

The concentration of the weak base solution = 0.500 m

The pH of the weak base = 9.75

Required:

The hydrolysis constant for the base

Solution:

Equation for the reaction is BOH $\longrightarrow$ B⁺(aq) + OH⁻(aq)

pH = 9.75

∴ [H⁺] ≈ 1.778 × 10⁻¹⁰

The pOH = 14 - 9.75 = 4.25

pOH = -log[OH⁻]

Therefore;

4.25 =  -log₁₀[OH⁻]

[OH⁻] = $10^{-4.25}$ = 5.623 × 10⁻⁵ moles

Number of moles of BOH that dissociates in the solution is therefore;

OH⁻ = B⁺ = 5.623 × 10⁻⁵ moles

The number of moles of BOH left in the solution is given as follows;

BOH = (0.5 m - 5.623 × 10⁻⁵ )

The hydrolysis constant, $k_h$, is given as follows;

$k_h = \dfrac{[Acid][Base]}{[Salt]}$

$k_h = \dfrac{[H^+][BOH]}{[B^+]}$

$k_h = \dfrac{1.778\times 10^{-10} \times (0.5 - 5.623 \times 10^{-5} ) }{5.623 \times 10^{-5}} \approx 1.581\times 10^{-6}$

The hydrolysis constant, $k_h$ ≈ 1.581 × 10⁻⁶

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