Hi laney,
For this question, it would be appropriate for you to ask yourself.. what is present in the solution after mixing?
KOH is a strong basic salt while HCL is a strong acidic salt.
KOH + HCL --> KCL + H2O
Note that amount of KOH = HCL = 0.125 X (50/1000), so there is no limiting reagent which means that both KOH and HCL are completely used up. What's left in the solution in this case - is just KCL (and H2O).
This being an acid-base reaction of a strong acid and strong base, will form a neutral salt KCL. Since the salt and H2O are neutral, pH of the solution would be 7.00.
In this case, because both reactants are strong acids/bases, no calculation is needed since a neutral salt is produced.
Hope I helped! (:
-J
The pH of a solution prepared by mixing 50.0 mL of 0.125 M KOH and 50.0 mL of 0.125 M HCl is __________.
i know the answer is 7.00, but im not sure how to solve it, do i create an ice chart?
2 answers
If necessary, you CAN calculate the pH at the neutral point.
HOH ==> H^+ + OH^-
Kw = (H^+)(OH^-) = 1 x 10^-14
(H^+) = (OH^-); therefore,
(H^+)^2 = 1 x 10^-14 and
(H^+) = 1 x 10^-7 M
pH = -log(H^+) = 7.0
HOH ==> H^+ + OH^-
Kw = (H^+)(OH^-) = 1 x 10^-14
(H^+) = (OH^-); therefore,
(H^+)^2 = 1 x 10^-14 and
(H^+) = 1 x 10^-7 M
pH = -log(H^+) = 7.0