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!
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)
is Kw 1.0e-14?
yes
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?
Yes, you solved both problems correctly! Your approach involved using the equation for the dissociation of a weak acid (or base) and the expression for the acid dissociation constant (Ka) to determine the concentration of the dissociated species, which you then used to calculate the pH.
In the first problem, you correctly set up the equation for the dissociation of acetic acid (CH3COOH): CH3COOH ⇌ CH3COO- + H+. The Ka value given is 1.8e-5, which represents the equilibrium constant for this reaction. Assuming that the concentration of H+ produced from the dissociation of acetic acid is x, you used the equation x^2/3.18 = 1.8e-5, since the initial concentration of acetic acid is 3.18M. Solving this equation, you found x to be 7.56e-3, which represents the concentration of H+ ions in the solution. Finally, you calculated the pH as -log(7.56e-3) = 2.12.
In the second problem, you applied the same logic. Here, you correctly set up the equation for the dissociation of hydrogen cyanide (HCN): HCN ⇌ CN- + H+. The Ka value given is 4.0e-10, which represents the equilibrium constant for this reaction. Assuming that the concentration of H+ produced from the dissociation of HCN is x, you used the equation x^2/1.75 = 4.0e-10, since the initial concentration of NaCN (which would dissociate to form HCN) is 1.75M. Solving this equation, you found x to be 2.64e-5, which represents the concentration of H+ ions in the solution. Finally, you calculated the pH as -log(2.64e-5) = 4.58.
Overall, your approach and calculations are correct for both problems. Well done!