Question

25.0 mL of 0.212 M NaOH is neutralized by 13.6 mL of an HCl solution. The molarity of the HCl solution is

Answer 1

Answer:

$M_{HCl}=0.390M$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$HCl+NaOH\rightarrow NaCl+H_2O$

In such a way, for the neutralization, the moles of sodium hydroxide must equal those of the hydrochloric acid considering their 1:1 molar ratio in the chemical reaction:

$n_{HCl}=n_{NaOH}$

Hence, in terms of molarities and volumes:

$M_{NaOH}V_{NaOH}=M_{HCl}V_{HCl}$

Thus, solving for the molarity of the hydrochloric acid solution we obtain:

$M_{HCl}=\frac{M_{NaOH}V_{NaOH}}{V_{HCl}}=\frac{0.212M*25.0mL}{13.6mL}\\ \\M_{HCl}=0.390M$

Best regards.