Posted by Hannah on Monday, April 2, 2012 at 3:47pm.
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
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
- Chemistry(Please check) - DrBob222, Monday, April 2, 2012 at 6:50pm
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.
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