Answer:
pH = 12.08
Explanation:
First we <u>calculate how many moles of each substance were added</u>, using <em>the given volume and concentration</em>:
- 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 <u>calculate how many OH⁻ moles remain after the reaction</u>:
- 5.55 - 3.75 = 1.8 mmoles OH⁻
With that<em> number of moles and the volume of the mixture</em>, we <u>calculate [OH⁻]</u>:
- [OH⁻] = 1.8 mmol / (75 mL + 74 mL) = 0.0121 M
With [OH⁻], we <u>calculate the pOH</u>:
With the pOH, we <u>calculate the pH</u>:
What are the answer choices ?
The charge of the cadmium cation is +2
