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CH3COOH--->CH3COO- + H+

if the freezing point of a .106 m CH3COOH solution is -.203 celcius, calculate the percent of the acid that has undergone ionization.

here's what i did, i took .203 celcius=i)1.86 celcius/m (.106m)

then i put .203/1.86(.106)=.103 which is i

then i took .103/2*100=5.15%

2 is the actual number formula units dissolved in the solution,

so did i do it right, it's a weak acid right, but i'm not sure if the percent is right