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valentina_108 [34]
2 years ago
14

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 t

he 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.
Chemistry
1 answer:
inn [45]2 years ago
7 0

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.

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Answer:

Concentration of the resulting solution = 2.04 M

Explanation:

<u>Data:</u>

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