Question

Formic acid $\left(HC{O}_{2}H\right)$ is a monoprotic acid that ionizes only partially in aqueous solutions. A 0.10-M formic acid solution is 4.2% ionized. Assuming that the molarity and molality of the solution are the same, calculate the freezing point and the boiling point of 0.10 M formic acid.

Short answer

Freezing point depression is $-0.19°C$and the boiling point elevation is $0.053°C$.

Step-by-step solution

Step 1:

Freezing point depression is $\Delta T_f=i\times K_f\times m$.

$\Delta T=1.042\times 1.86°Ckg/mo{l}^{-1}\times 0.10mol/kg=0.19°C$

Step 2:

The boiling point elevation is $\Delta T_f=i\times K_b\times m$

$\Delta T=1.042\times 0.512°Ckg/mo{l}^{-1}\times 0.10mol/kg=0.053°C$