Question

The reaction of HCl with NaOH is represented by the equation HCl( aq) + NaOH( aq) ® NaCl( aq) + H 2O( l) What volume of 0.631 M HCl is required to titrate 15.8 mL of 0.321 M NaOH?

Answer 1

8.04 mL.

Explanation

How many moles of NaOH?

Note the unit:

$V = 15.8 \; \textbf{mL} = \dfrac{15.8}{1000}\;\textbf{L} = 0.0158 \; \text{L}$.

$n = c\cdot V = 0.0158 \times0.321 = 0.00507\;\text{mol}$.

How many moles of HCl?

As seen in the equation, HCl and NaOH reacts at a 1:1 ratio.

$n(\text{HCl}) = n(\text{NaOH}) = 0.00507\;\text{mol}$.

How many mL of HCl?

$V = \dfrac{n}{c} =\dfrac{0.00507}{0.631} = 0.00804\;\text{L} = 0.00804\times 1000 \;\textbf{mL} = 8.04\;\text{mL}$.