Answer to Question #94663 in General Chemistry for Olissa

Question #94663
A 0.105m HF (aq) solution shows a freezing point of -0.198 C. Calculate the % dissociation of HF.
1
Expert's answer
2019-09-19T03:58:45-0400

The freezing point depression equation is:


ΔTf=i×Kf×m\Delta T_f= i\times K_f\times m

The only value we do not know is i:


0.198=i×1.86×0.1050.198 = i\times 1.86 \times 0.105


i=1.014i = 1.014


% ionozation =(i1)×100%= (i-1)\times 100 \%


% ionization = (i-1)\times 100% % ionization = =(1.0141)×100%=14%=(1.014-1)\times 100 \% = 14\%

% ionization = (i-1)\times 100%

The HF is about 14% dissociated.


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