Posted by Hannah on .
1) The Ka for acetic acid is 1.8e5. What is the pH of a 3.18M solution of this acid?
I did 1.8e5 = x^2/3.18
x=sqrt 1.8e5 X 3.18 = 7.56e3
pH=log(7.56e3) = 2.12
The pH is 2.12
2) At 25 degrees celsius, the pH of a 1.75M NaCN solution (the Ka for HCN is 4.0e10) is ?
4.0e10 = x^2/1.75
x=sqrt 4.0e10 X 1.75 = 2.64e5
pH=log(2.64e5) = 4.58
Did I solve these problems correctly? Thank you!
Chemistry(Please check answers)  DrBob222, Sunday, April 1, 2012 at 6:03pm
1 is right.
2 is not.
The pH of salts is determined by the hydrolysis of the salt.
CN^ + HOH ==> HCN + OH^
Kb for CN^ is (Kw/Ka for HCN)
Chemistry(Please check answers)  Hannah, Sunday, April 1, 2012 at 9:33pm
is Kw 1.0e14?
Chemistry(Please check answers)  DrBob222, Sunday, April 1, 2012 at 9:36pm
yes
Chemistry(Please check answers)  Hannah, Sunday, April 1, 2012 at 9:47pm
so kb = 1.0e14 / 4.0e10 = 2.5e5
Is 2.5e5 my answer or do I have to do another step

Chemistry(Please check) 
DrBob222,
You have calculated Kb correctly. Now you need to substitute that into the other part of the equation. You previously had this:
4.0e10 = x^2/1.75M.
Everything is ok except the K value. You had the k value for HCN, you need the Kb value for the CN^ which you have now calculated to be 2.5E5.
So 2.5E5 = (x)^2/1.75 and solve for x = OH^ then convert to pH.