Question

A 0.400M formic acid (HCOOH) solution freezes at −0.758∘C Calculate the Ka of the acid at that temperature. (Hint: Assume that molarity is equal to molality. Carry out your calculations to three significant figures and round off to two for Ka.)

Answer 1

First, find the molality of the solution.

`i=(T_f)/(k_f\cdot m)`

Where k is the molal freezing point, which is 1.86 C/m for water.

`i=(0.758)/(1.86\cdot0.4)=(0.758)/(0.744)=1.02`

The equilibrium constant Ka would be

`K_a=(\lbrack HCOO\rbrack\lbrack H\rbrack)/(\lbrack HCOOH\rbrack)`
`K_a=(x\cdot x)/(0.4-x)`

Then, to find x.

`\begin{gathered} i=(0.4-x+x+x)/(0.4)=1.02 \\ x=1.02\cdot0.4-0.4=0.008 \end{gathered}`

Once we have x, we can obtain the constant Ka

`K_a=((0.008)^2)/(0.4-0.008)=(6.4*10^(-5))/(0.392)=1.6*10^(-4)`

Therefore, the constant Ka of the reaction is 1.6x10^-4.