What is the pH of a solution made by mixing 30.00 mL 0.10 M HCl with 40.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive.

pH = 12.15

Explanation:

To determine the pH of the HCl and KOH mixture, we need to know that the reaction is  a neutralization type.

HCl  +  KOH  →  H₂O  +  KCl

We need to determine the moles of each compound

M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl

M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH

The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl

HCl  +  KOH  →  H₂O  +  KCl

3 m       4 m                       -

             1 m                      3 m

As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.

1 mmol of KOH has 1 mmol of OH⁻, so the [OH⁻] will be 1 mmol / Tot volume

[OH⁻] 1 mmol / 70 mL = 0.014285 M

- log [OH⁻] = 1.85 → pH = 14 - pOH → 14 - 1.85 = 12.15

thats how its supposed to be