Question

What is the molar concentration of H2SO4 in a solution made by reacting 188.9 mL of 2.086 M H2SO4 with 269.3 mL of 0.4607 M NaOH?

Answer 1

Answer:

0.7246 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $H_2SO_4$ :

Molarity = 2.086 M

Volume = 188.9 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 188.9×10⁻³ L

Thus, moles of $H_2SO_4$ :

$Moles=2.086 \times {188.9\times 10^{-3}}\ moles$

Moles of $H_2SO_4$  = 0.39405 moles

For NaOH :

Molarity = 0.4607 M

Volume = 269.3 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 269.3×10⁻³ L

Thus, moles of NaOH :

$Moles=0.4607 \times {269.3\times 10^{-3}}\ moles$

Moles of NaOH  = 0.1241 moles

According to the given reaction:

$H_2SO_4_{(aq)}+2NaOH_{(aq)}\rightarrow Na_2SO_4_{(aq)}+2H_2O_{(aq)}$

1 moles of $H_2SO_4$ react with 2 moles of NaOH to form 1 mole of sodium sulfate.

Thus,

2 moles of NaOH react with 1 mole of $H_2SO_4$

1 mole of NaOH react with 1/2 mole of $H_2SO_4$

0.1241 moles of NaOH react with (1/2)×0.1241 mole of $H_2SO_4$

Moles of $H_2SO_4$ that got reacted = 0.06205 moles

Unreacted moles = Total moles - Moles that got reacted = 0.39405 - 0.06205 moles = 0.332 moles

Total volume = 188.9×10⁻³ L + 269.3×10⁻³ L = 458.2×10⁻³ L

Concentration of $H_2SO_4$ :

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{H_2SO_4}=\frac{0.332}{458.2\times 10^{-3}}$

Concentration of $H_2SO_4$ = 0.7246 M