Answer to Question #94663 in General Chemistry for Olissa

Question

A 0.105m HF (aq) solution shows a freezing point of -0.198 C. Calculate the % dissociation of HF.

Answer 1

The freezing point depression equation is:

$$\Delta T_f= i\times K_f\times m$$

The only value we do not know is i:

$$0.198 = i\times 1.86 \times 0.105$$

$$i = 1.014$$

% ionozation $= (i-1)\times 100 \%$

$\%\; ionization\; =\; (i-1)\backslash times\; 100\%$ % ionization = $=(1.014-1)\times 100 \% = 14\%$

$$\%\; ionization\; =\; (i-1)\backslash times\; 100\%$$

The HF is about 14% dissociated.