Question

A solution containing 0.739 M KOH completely neutralizes 12.0 mL of 0.550 M H2SO4. What volume of KOH was used? 

Answer 1

To solve this question, we would use mole concept

step one

write a balanced chemical equation

$H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O$

Step two

Using the statement given,

0.55 mol of H₂SO₄ = 1000mL

x mol = 12.0mL

$\begin{gathered} 0.55m\rightarrow1000ml \\ xm\rightarrow12.0ml \\ \text{cross multiply both sides and solve for x} \\ x=\frac{12\times0.55}{1000} \\ x=0.0066\text{moles} \end{gathered}$

Going back to the balanced chemical equation,

1 mole of H₂SO₄ reacted with 2 moles of KOH

This implies that 0.0066 moles of H₂SO₄ will react to y

Cross multiply both sides and solve for y.

$\begin{gathered} 1\text{mol}H_{2}SO_4\rightarrow2molKOH \\ 0.0066\text{mol}\rightarrow y \\ y=0.0066\times2 \\ y=0.0132\text{mol} \end{gathered}$

Now we can finally proceed to find the volume of KOH used.

If 0.739moles is present in 1000mL of KOH

0.0132 moles is present in z mL

$\begin{gathered} 0.739mol\rightarrow1000mL \\ 0.0132\text{mol}\rightarrow zmol \\ z=\frac{0.0132\times1000}{0.739} \\ z=17.86mL \end{gathered}$

From the calculation above, 17.86mL of KOh reacted with 12.0mL of H₂SO₄