Question

A 0.1 molal solution of a weak monoprotic acid was found to depress the freezing point of water 0.1930C. Determine the Ka of the acid. You can assume 0.1 molal and 0.1 molar are equivalent.


100 ml of a 0.1M solution of the above acid is titrated with 0.1 M NaOH. After an amount of NaOH is added the pH= 5.2. How much NaOH was added?


( c ) The titration is continued to equivalence. What is the pH at equivalence?

Answer 1

I'll get you started by helping with the first part.
delta T = Kf*m
0.1930 = 1.86*m
m = 0.1930/1.86 = 0.10376 (Since the delta T is to 4 significant figures, I've use more places in my answer. I don't know if that is 0.1M or 0.100 M so watch the number of s.f. when you do this yourself.)
Since the solution is 0.1, that means the solution actually contains more particles than 1 (it is ionized slightly).
0.10376-0.1 = 0.0376 extra particles
%ion = (0.00376/0.1)*100 = about 4%
.............HA ==> H^+ + A^-
initial..0.100..0.00376..0.00376
change.....-x........x......x
equil....0.1-0.00376

Ka = set up Ka expression and substitute values from the ICE chart. Solve for Ka.