Question

Determine the resulting pH when 75 mL of 0.05M HBr is reacted with 74 mL of 0.075 M KOH

Answer 1

Answer:

pH = 12.08

Explanation:

• H⁺ + OH⁻ → H₂O

First we calculate how many moles of each substance were added, using the given volume and concentration:

• HBr ⇒ 0.05 M * 75 mL = 3.75 mmol HBr

• KOH ⇒ 0.075 M * 74 mL = 5.55 mmol KOH

As HBr is a strong acid, it dissociates completely into H⁺ and Br⁻ species. Conversely, KOH dissociates completely into OH⁻ and K⁺ species.

As there are more OH⁻ moles than H⁺ moles (5.55 vs 3.75), we calculate how many OH⁻ moles remain after the reaction:

• 5.55 - 3.75 = 1.8 mmoles OH⁻

With that number of moles and the volume of the mixture, we calculate [OH⁻]:

• [OH⁻] = 1.8 mmol / (75 mL + 74 mL) = 0.0121 M

With [OH⁻], we calculate the pOH:

• pOH = -log[OH⁻] = 1.92

With the pOH, we calculate the pH:

• pH = 14 - pOH

• pH = 12.08