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Answer to Question #6521 in Chemistry for Dee
Question #6521
if a weak acid, HA, is 3% dissociated in a 0.25M solution, calculate the Ka and the pH of the solution.
Expert's answer
a=3% (0.03)
C=0.25M
Ka-? pH-?
HA <=>H+ +
A-
a=(K/C) ^ (1/2)
a^ 2= K/C
K=(a^ 2)/C
K= (0.03^ 2)
/0.25=0.0036
[H+]=(K*C) ^ (1/2)=0.03
pH=-log[H+]=1.52
K=0.0036
pH=1.52
C=0.25M
Ka-? pH-?
HA <=>H+ +
A-
a=(K/C) ^ (1/2)
a^ 2= K/C
K=(a^ 2)/C
K= (0.03^ 2)
/0.25=0.0036
[H+]=(K*C) ^ (1/2)=0.03
pH=-log[H+]=1.52
K=0.0036
pH=1.52
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Comments
The % dissociated (or value "a") tells you that you have 3% of A- and 97% of HA. For a 0.25M solution, that means you have (0.97 * 0.25)M HA present, (0.03 * 0.25)M A- present, and (0.03 * 0.25)M H+ present. Putting these concentrations into the equation doesn't give you K = 0.0036. In fact, the first problem is that whoever answered before assumes "% dissociated" is the concentration of H+ present...a mistake which can be seen in solving the second part.
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