Question

If the titration of 25.0 mL sample of hydrochloric acid (HCl) requires 22.15 mL of 0.155M sodium hydroxide (NaOH), what is the molarity of the acid?

Answer 1

Given:

Volume of HCl (VHCl) = 25,0 ml = 0.025 L

Volume of NaOH (VNaOH) = 22.15 ml = 0.02215 L

Molarity of NaOH (MNaOH) = 0.155 M

To determine:

Molarity of HCl

Explanation:

Titration reaction-

HCl + NaOH →  NaCl + H2O

Based on the reaction stoichiometry:

1 mole NaOH requires 1 mole HCl

# moles of NaOH = VNaOH * MNaOH = 0.02215*0.155 = 0.003433 moles

# moles of HCl = # moles of NaOH = 0.003433 moles

Molarity of NaOH = 0.003433/0.025 = 0.137 M

Answer 2

The molarity of the acid is 0.137 M

From the question,

We are to determine the molarity (that is, concentration) of the acid,

First, we will write a balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

HCl + NaOH → NaCl + H₂O

Now, to determine the molarity of the acid,

We will use the formula

$\frac{C_{A}V_{A} }{C_{B}V_{B}} = \frac{n_{A}}{n_{A}}$

Where

$C_{A}$ is the concentration of acid

$C_{B}$ is the concentration of base

$V_{A}$ is the volume of acid

$V_{B}$ is the volume of base  

$n_{A}$ is the mole ratio of acid

$n_{B}$ is the mole ratio of base

From the question

$C_{B} = 0.155 \ M$

$V_{A} = 25.0 \ mL$

$V_{B} = 22.15 \ mL$

From the balanced chemical equation

$n_{A} = 1$

$n_{B} = 1$

Putting the above parameters into the formula

$\frac{C_{A}V_{A} }{C_{B}V_{B}} = \frac{n_{A}}{n_{A}}$

We get

$\frac{C_{A} \times 25.0}{0.155 \times 22.15} = \frac{1}{1}$

Then,

$C_{A} \times 25.0 = 0.155 \times 22.15$

∴ $C_{A} = \frac{0.155 \times 22.15}{25.0}$

$C_{A} = \frac{3.43325}{25.0}$

$C_{A} = 0.13733 \ M$

$C_{A} \approx 0.137 \ M$

Hence, the molarity of the acid is 0.137 M

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