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

0.7246 M

Step-by-step explanation:

Considering:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

Or,

`Moles =Molarity * {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 * {188.9* 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 * {269.3* 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=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

`Molarity_(H_2SO_4)=(0.332)/(458.2* 10^(-3))`

Concentration of
`H_2SO_4` = 0.7246 M