Problem: Find the percent ionization of a 0.337 M HF solution. The Ka for HF is 3.5 x 10-4. A) 4.7% B) 3.2% C) 1.2 x 10-2% D) 3.5 x 10-2%E) 1.1 % 

FREE Expert Solution

% ionization=[H+][HA]initial×100



HF(aq) + H2O(l)  F(aq) + H3O+(aq); Ka = 3.5 × 10–4


From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table and Ka expression.



The Ka expression for HF is:


Ka=productsreactants=[H3O+][F-][HF]



Plugging in the equilibrium concentrations from the ICE table into the Ka expression:


Ka=3.5×10-4=[H3O+][F-][HF]


Ka=3.5×10-4=(x)(x)0.337-x


Now, we need to determine if we can remove (–x) from the equation. To do so, we need to determine the ratio of the initial concentration and Ka:


[HF]initialKa>500


Since the ratio is greater than 500, we can remove (–x) from the equation. 

The equation simplifies to:

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Problem Details

Find the percent ionization of a 0.337 M HF solution. The Ka for HF is 3.5 x 10-4

A) 4.7% 

B) 3.2% 

C) 1.2 x 10-2

D) 3.5 x 10-2%

E) 1.1 % 

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