Question

What volume of a 0.317 M HBR solution is required to neutralize 18.8mL of a 0.129 M KOH solution?

Answer 1

Answer:

$\large \boxed{\text{7.65 mL}}$

Explanation:

1. Balanced chemical equation.

$\rm HBr + KOH \longrightarrow KBr + H_{2}O$

2. Moles of NaOH

$\text{Moles of NaOH} =\text{ 18.8 mL NaOH} \times \dfrac{\text{0.129 mmol NaOH }}{\text{1 mL NaOH }} = \text{2.425 mmol NaOH }$

3. Moles of HBr

The molar ratio is 1 mmol HBr:1 mmol NaOH

$\text{Moles of HBr}= \text{2.425 mmol NaOH} \times \dfrac{\text{1 mmol HBr}}{\text{1 mmol NaOH}} =\text{2.425 mmol HBr}$

4. Volume of HBr

$V = \text{2.425 mmol HBr}\times \dfrac{\text{1mL HBr}}{\text{0.317 mmol HBr}} = \textbf{7.65 mL HBr}\\\\\text{The reaction requires \large \boxed{\textbf{7.65 mL HBr}} }$