You need the molarity.
pH = -log(H^+)
Calculate the pH of a [Acid] M aqueous solution of each of the hydrochloric and acetic acids:
Acid [Acid] pH
HCl 0.10
HC2H3O2 0.10
I know that HCl is just -log(0.10) which is 1, but can't get the pH of HC2H3O2.
4 answers
1, 2,87
i'm doing this problem too =)
Molarity for both hydrochloric and acetic acids are 0.10. For hydrochloric acid, i used the formula: pH = -log (0.10) = 1.
But should we do something else for acetic acid? because its pH is not 1.
Molarity for both hydrochloric and acetic acids are 0.10. For hydrochloric acid, i used the formula: pH = -log (0.10) = 1.
But should we do something else for acetic acid? because its pH is not 1.
never mind, I found out how to do this =)
(the post above was me by the way).
set up I.C.E. table for the equilibrium expression. Since we know that Ka value for acetic acid is about 1.76 x 10^-5, we can solve for x (which is [H+] in this case).
Ka = x^2 / 0.10 - x => approximation: 1.76 x 10^-5 = X^2 / 0.10. Solve for x. Then use pH = -log [H+] to find pH of acetic acid =).
(the post above was me by the way).
set up I.C.E. table for the equilibrium expression. Since we know that Ka value for acetic acid is about 1.76 x 10^-5, we can solve for x (which is [H+] in this case).
Ka = x^2 / 0.10 - x => approximation: 1.76 x 10^-5 = X^2 / 0.10. Solve for x. Then use pH = -log [H+] to find pH of acetic acid =).
1 answer