Answer:
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
We know that [OH⁻] * [H⁺] = 10⁻¹⁴
plugging the value of [H⁺]
[OH⁻] * 1.2 * 10⁻³ = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴ * (10³/1.2)
[OH⁻] = 833.3 * 10⁻¹⁴
[OH⁻] = 8.33 * 10⁻¹²
