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

`M_(HCl)=7.96M`

Step-by-step 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)=(M_(NaOH)V_(NaOH))/(V_(HCl)) =(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.