# Question 0163b

Mar 7, 2015

The molarity of the potassium hydroxide solution is $\text{0.951 M}$.

So, you've got your balanced chemical equation for what is essentially a neutralization reaction

$2 K O {H}_{\left(a q\right)} + {H}_{2} S {O}_{4 \left(a q\right)} \to {K}_{2} S {O}_{4 \left(a q\right)} + 2 {H}_{2} {O}_{\left(l\right)}$

When you performed your titration and reached the equivalence point, you essentially added enough sulfuric acid to completely react with the potassium hydroxide solution, i.e. you've added enough protons to react will all the hydroxide ions and form water.

The volume and the concentration of the sulfuric acid solution will help you determine how many moles were needed to completely react with the potassium hydroxide.

$C = \frac{n}{V} \implies {n}_{{H}_{2} S {O}_{4}} = C \cdot V$

${n}_{{H}_{2} S {O}_{4}} = \text{1.50 M" * 22.2*10^(-3)"L" = "0.0333 moles}$

Now look at the mole ratio that exists between $K O H$ and ${H}_{2} S {O}_{4}$. Notice that, for every 1 mole of the latter, you need 2 moles of the former for the reaction to take place.

This means that, regardless of how many ${H}_{2} S {O}_{4}$ moles you have, you'll always need twice as many $K O H$ moles present in solution. Since you know how many moles you've added, you can determine how many moles of $K O H$ were present in the solution by

$\text{0.0333 moles sulfuric acid" * "2 moles potassium hydroxide"/"1 mole sulfuric acid" = "0.0666 moles KOH}$

SInce you had this many moles in your initial 70.0 mL volume, the concentration can only be

C_("KOH") = n/V = "0.0666 moles"/(70.0 * 10^(-3)"L") = "0.951 M"#