Question

Calculate the OH− concentration after 53 mL of the 0.100 M KOH has been added to 25.0 mL of 0.200 M HBr. Assume additive vol- umes.

Answer in units of M.

Answer 1

`\large \boxed{\text{0.0038 mol/L}}`

Step-by-step explanation:

1. Calculate the initial moles of acid and base

`\text{moles of acid} = \text{0.0250 L} * \frac{\text{0.200 mol}}{\text{1 L}} = \text{0.005 00 mol}\\\\\text{moles of base} = \text{0.053 L} * \frac{\text{0.100 mol}}{\text{1 L}} = \text{0.0053 mol}`

2. Calculate the moles remaining after the reaction

OH⁻ + H₃O⁺ ⟶ 2H₂O

I/mol: 0.0053 0.005 00

C/mol: -0.00500 -0.005 00

E/mol: 0.0003 0

We have an excess of 0.0003 mol of base.

3. Calculate the concentration of OH⁻

Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L

`\text{[OH}^(-)] = \frac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^(-)$ is $\large \boxed{\textbf{0.0038 mol/L}}$}`