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Question: If degree of dissociation is 0.01 of decimolar solution of weak acid HA then $pK_a$ of acid is : 41....

If degree of dissociation is 0.01 of decimolar solution of weak acid HA then pKapK_a of acid is : 41. What concentration of HCOO-

A

2

B

3

C

5

D

7

Answer

5

Explanation

Solution

The dissociation of a weak acid HA is given by the equilibrium:

HA⇌H++A−HA \rightleftharpoons H^+ + A^-

Let CC be the initial concentration of the weak acid and α\alpha be its degree of dissociation. At equilibrium, the concentrations are:

[HA]=C(1−α)[HA] = C(1-\alpha)
[H+]=Cα[H^+] = C\alpha
[A−]=Cα[A^-] = C\alpha

The acid dissociation constant KaK_a is given by:

Ka=[H+][A−][HA]=(Cα)(Cα)C(1−α)=Cα21−αK_a = \frac{[H^+][A^-]}{[HA]} = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}

We are given that the solution is decimolar, which means the concentration C=0.1C = 0.1 M. The degree of dissociation is given as α=0.01\alpha = 0.01.

Since the degree of dissociation α=0.01\alpha = 0.01 is much less than 1, we can use the approximation 1−α≈11-\alpha \approx 1. Using this approximation, the formula for KaK_a becomes:

Ka≈Cα2K_a \approx C\alpha^2

Substitute the given values of CC and α\alpha:

Ka=(0.1)×(0.01)2K_a = (0.1) \times (0.01)^2
Ka=0.1×(10−2)2K_a = 0.1 \times (10^{-2})^2
Ka=0.1×10−4K_a = 0.1 \times 10^{-4}
Ka=10−1×10−4K_a = 10^{-1} \times 10^{-4}
Ka=10−5K_a = 10^{-5}

The pKapK_a of the acid is related to KaK_a by the formula:

pKa=−log⁡10(Ka)pK_a = -\log_{10}(K_a)

Substitute the calculated value of KaK_a:

pKa=−log⁡10(10−5)pK_a = -\log_{10}(10^{-5})

Using the property of logarithms log⁡10(xy)=ylog⁡10(x)\log_{10}(x^y) = y\log_{10}(x):

pKa=−(−5×log⁡10(10))pK_a = -(-5 \times \log_{10}(10))

Since log⁡10(10)=1\log_{10}(10) = 1:

pKa=−(−5×1)pK_a = -(-5 \times 1)
pKa=5pK_a = 5

Therefore, the pKapK_a of the acid is 5.