Question

Concentrated hydrochloric acid is a solution that is 37.5% mass per unit volume HCl(aq) in water. An old bottle of HCl has an unknown concentration. What is the concentration of hydrochloric acid, [HCl], in the old bottle, if 9.95 mL of 12.0 M NaOH(aq) is required to reach the equivalence point when added to 15 mL of acid?What is the concentration of HCl(aq)?

Answer 1

Answer:

$M_{HCl}=7.96M$

Explanation:

Hello,

In this case, since the neutralization reaction between HCl and NaOH is:

$HCl+NaOH\rightarrow NaCl+H_2O$

We notice a 1:1 molar ratio, for that reason, at the equivalence point we find:

$n_{HCl}=n_{NaOH}$

Thus in terms in molarities one could compute the concentration of HCl in the old bottle for the used NaOH for the neutralization as:

$M_{HCl}V_{HCl}=M_{NaOH}V_{NaOH}\\\\M_{HCl}=\frac{M_{NaOH}V_{NaOH}}{V_{HCl}} =\frac{12.0M*9.95mL}{15mL}\\ \\M_{HCl}=7.96M$

This value is lower than 37% HCl that in molarity is about 12 M, such difference is due to its high volatility.

Best regards.