Question

student titrated 15.00 mL of HCl of an unknown concentration with a solution of 0.0670 M NaOH. This titration used 19.06 mL of the 0.0670 M NaOH to reach the end point. Determine the concentration of the unknown HCl in one set up. You may look at the steps used in the previous problem to recall what fractions need to be in the set-up to solve this problem. 1. Fill in the fractions required to determine the concentration of HCl.

Answer 1

Answer:

The concentration of the unknown HCl is 0.0851 M.

Explanation:

The equation of neutralization:

$n_1M_1V_1=n_2M_2V_2$

Where:

$n_1$= Basicity of acid

$n_2$ = Acidity of base

$M_1$= concentration  of acid

$M_2$= concentration  of base

$V_1$ = Volume of acid used in neutralization

$V_2$ = Volume of base used in neutralization

We have:

The acidity of HCl = $n_1=1$

The concentration of HCl solution used = $M_1=?$

The volume of HCl used in titration =$V_1= 15.00 mL$

The acidity of NaOH =$n_2=1$

The concentration of NaOH solution used = $M_2=0.0670 M$

The volume of NaOH used in titration =$V_2= 19.06 mL$

$n_1M_1V_1=n_2M_2V_2\\1\times M_1\times 15.00 mL=1\times 0.0670 M\times 19.06 mL\\M_1=\frac{1\times 0.0670 M\times 19.06 mL}{1\tines 15.00 mL}\\M_1=0.085135 M\approx 0.0851 M$

The concentration of the unknown HCl is 0.0851 M.