Chemistry Daily Challenge 21-July-2015

Chemistry Level 3

Suppose there is a 0.01 M solution of a weak monobasic acid whose acid dissociation constant is $\text{p}K_a=4.$ What is the van't Hoff factor of this solution?

The answer is 1.1.

1 solution

Chew-Seong Cheong
Jul 24, 2015

For monobasic acid: $HA \longrightarrow H^+ + A^-$ . Let concentration be $a= 0.01\text{M}$ and degree of disassociation be $\alpha$ . Then, we have:

$\begin{aligned} K_a & = \frac{[H^+][A^-]}{[HA]} = \frac{(a\alpha)(a \alpha)}{a(1-\alpha)} = \frac{a\alpha^2}{1-\alpha} \end{aligned}$

For small $\alpha$ , we have $\alpha \approx \sqrt{\dfrac{K_a}{a}} = \sqrt{\dfrac{10^{-pK_a}}{a}} = \sqrt{\dfrac{10^{-4}}{0.01}} = 0.1$

The Van't hoff factor $i = \dfrac{1-\alpha+\alpha+\alpha}{1} = 1 + \alpha = 1+0.1 = \boxed{1.1}$

Moderator note:

Do you know why the approximation is valid? IE Why do we have

$\frac{ \alpha^2}{ 1 - \alpha } = \frac{ K_a } { a} \Rightarrow \alpha \approx \sqrt{ \frac{ K_a}{a} }$

You forgot to write -1.Any way nice explanation.

$\alpha \approx \sqrt{\dfrac{K_a}{a}} = \sqrt{\dfrac{10^{\color{#D61F06}{-pK_a}}}{a}} = \sqrt{\dfrac{10^{-4}}{0.01}} = 0.1$

Aamir Faisal Ansari - 5 years, 10 months ago

Thanks. I am learning this for the first time. I was referring to this video lecture van't Hoff Factor_Solutions by Er. Dushyant Kumar (B.Tech. IIT-Roorkee) . I could figure it out even though I don't know Hindi.

Chew-Seong Cheong - 5 years, 10 months ago

Great...happy to see that these sources are really helpful

Aamir Faisal Ansari - 5 years, 10 months ago

Grt!!!I respect your thirst for learning

Spandan Senapati - 4 years, 3 months ago

Because $\alpha \ll 1\quad \Rightarrow 1-\alpha \approx 1$ .

Chew-Seong Cheong - 5 years, 2 months ago

Sir it is said that if $pK_{a} \geq 5$ then only we can use that approx. So the answer is a little distant from correct one.

Md Zuhair - 3 years, 6 months ago

Chemistry Daily Challenge 21-July-2015

The answer is 1.1.

1 solution

Moderator note:

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