# 0.850 L of 0.490 M H_2SO_4 is mixed with 0.800 L of 0.280 M KOH. What concentration of H_2SO_4 remains after neutralization?

##### 1 Answer
Apr 8, 2016

The concentration of ${\text{H"_2"SO}}_{4}$ after partial neutralization is 0.185 mol/L

#### Explanation:

This is really a two-part question:

1. How much of the ${\text{H"_2"SO}}_{4}$ is neutralized?
2. What is the concentration of the remaining ${\text{H"_2"SO}}_{4}$?

1. How much of the ${\text{H"_2"SO}}_{4}$ is neutralized?

This is a volume-moles stoichiometry problem.

The steps are:

a. Write the balanced equation.

b. Use the molarity of $\text{KOH}$ to convert volume of $\text{KOH}$ to moles of $\text{KOH}$.

c. Use the molar ratio from the balanced equation to convert moles of $\text{KOH}$ to moles of ${\text{H"_2"SO}}_{4}$ that reacted.

d. Use the molarity to convert the initial volume of ${\text{H"_2"SO}}_{4}$ to moles of ${\text{H"_2"SO}}_{4}$.

e. Calculate the moles of excess ${\text{H"_2"SO}}_{4}$.

a. Write the balanced equation:

$\text{2KOH" + "H"_2"SO"_4 → "K"_2"SO"_4 + "2H"_2"O}$

b. Calculate the moles of $\text{KOH}$.

$\text{Moles of KOH" = 0.800 color(red)(cancel(color(black)("L KOH"))) × "0.280 mol KOH"/(1 color(red)(cancel(color(black)("L KOH")))) = "0.224 mol KOH}$

c. Calculate the moles of ${\text{H"_2"SO}}_{4}$ that reacted

The molar ratio of $\text{H"_2"SO"_4:"KOH}$ is $\text{1 mol H"_2"SO"_4:"2 mol KOH}$.

${\text{Moles of H"_2"SO"_4 = 0.224 color(red)(cancel(color(black)("mol KOH"))) × (1 "mol H"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol KOH")))) = "0.112 mol H"_2"SO}}_{4}$

2. What is the concentration of the remaining ${\text{H"_2"SO}}_{4}$?

a. Calculate the initial moles of ${\text{H"_2"SO}}_{4}$.

$\text{Moles of H"_2"SO"_4 = 0.850 color(red)(cancel(color(black)("L H"_2"SO"_4))) × ("0.490 mol H"_2"SO"_4)/(1 color(red)(cancel(color(black)("L H"_2"SO"_4)))) = "0.4165 mol}$

e. Calculate the moles of ${\text{H"_2"SO}}_{4}$ remaining

$\text{Moles remaining" = "initial moles – moles reacted" = "0.4165 mol – 0.112 mol" = "0.3045 mol}$

e. Calculate the molarity of the excess ${\text{H"_2"SO}}_{4}$

$\text{Molarity" = "moles"/"litres}$

$\text{Litres" = "Volume of KOH + Volume of H"_2"SO"_4 = "0.800 L + 0.850 L" = "1.650 L}$

$\text{Molarity" = "0.3045 mol"/"1.650 L" = "0.185 mol/L}$