1) The Ka for acetic acid is 1.8e-5. What is the pH of a 3.18M solution of this acid?

I did 1.8e-5 = x^2/3.18

x=sqrt 1.8e-5 X 3.18 = 7.56e-3

pH=-log(7.56e-3) = 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 4.0e-10) is ?

4.0e-10 = x^2/1.75

x=sqrt 4.0e-10 X 1.75 = 2.64e-5

pH=-log(2.64e-5) = 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 1.0e-14?

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 = 1.0e-14 / 4.0e-10 = 2.5e-5

Is 2.5e-5 my answer or do I have to do another step

You have calculated Kb correctly. Now you need to substitute that into the other part of the equation. You previously had this:

4.0e-10 = 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 2.5E-5.
So 2.5E-5 = (x)^2/1.75 and solve for x = OH^- then convert to pH.

Yes, 2.5e-5 is your answer. Since you have correctly calculated the value of Kb for CN-, you don't need to do any further steps in the calculation. The pH of a solution can be determined using the expression:

pOH = -log(OH- concentration)
pH = 14 - pOH
In this case, the CN- ion undergoes hydrolysis and produces OH- ions, so the pOH can be found using the Kb value:
Kb = [OH-][HCN]/[CN-]
Since the concentration of CN- is 1.75M, and the concentration of OH- is the same as the concentration of HCN, you can solve for [OH-]:
[OH-] = sqrt(Kb * [CN-]) = sqrt(2.5e-5 * 1.75) = 7.47e-3
Finally, calculate the pH:
pOH = -log(7.47e-3) = 2.13
pH = 14 - 2.13 = 11.87
So the pH of the 1.75M NaCN solution is approximately 11.87.