Answer:
[tex]M_{HCl}=7.96M[/tex]
Explanation:
Hello,
In this case, since the neutralization reaction between HCl and NaOH is:
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
We notice a 1:1 molar ratio, for that reason, at the equivalence point we find:
[tex]n_{HCl}=n_{NaOH}[/tex]
Thus in terms in molarities one could compute the concentration of HCl in the old bottle for the used NaOH for the neutralization as:
[tex]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[/tex]
This value is lower than 37% HCl that in molarity is about 12 M, such difference is due to its high volatility.
Best regards.
