# Question c83dc

Jul 26, 2015

pH = $1$

#### Explanation:

You're dealing with a neutralization reaction between sulfuric acid, a strong acid, and potassium hydroxide, a strong base.

This reaction will produce potassium sulfate, a salt, and water. To get an idea about how this reaction works, write the balanced chemical equation first

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

The $1 : \textcolor{red}{2}$ mole ratio that exists between sulfuric acid and potassium hydroxide tells you that, in order to have a complete neutralization, you need twice as many moles of potassium hydroxide than of sulfuric acid.

Use the volumes and molarities of the two solutions to see how many moles you add together

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{{H}_{2} S {O}_{4}} = \text{0.2 M" * 50 * 10^(-3)"L" = "0.001 moles}$ ${H}_{2} S {O}_{4}$

and

${n}_{K O H} = \text{0.2 M" * 50 * 10^(-3)"L" = "0.001 moles}$ $K O H$

You have equal numbers of moles of each reactant, which means that potassium hydroxide will act as a limiting reagent. In other words, you have insufficient moles of potassium hydroxide to react will all the moles of sulfuric acid.

This means that the reaction will only consume

0.001cancel("moles"KOH) * ("1 mole "H_2SO_4)/(color(red)(2)cancel("moles"KOH)) = "0.0005 moles" ${H}_{2} S {O}_{4}$

The remaining sulfuric acid will be in excess

n_(H_2SO_4"excess") = 0.001 - 0.0005 = "0.0005 moles"

To determine the pH of the solution, you first need to determine the concentration of protons present.

Since each mole of sulfuric acid dissociates to produce two moles of protons, you'll have

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

0.0005cancel("moles"H_2SO_4) * ("2 moles "H^(+))/(1cancel("mole"H_2SO_4)) = "0.001 moles"# ${H}^{+}$

The total volume of the solution will be

${V}_{\text{total}} = {V}_{{H}_{2} S {O}_{4}} + {V}_{K O H}$

${V}_{\text{total" = 50 + 50 = "100 mL}}$

The molarity of the protons will be

$\left[{H}^{+}\right] = \text{0.001 moles"/(100 * 10^(-3)"L") = "0.1 M}$

The pH of the solution will thus be

$p {H}_{\text{sol}} = - \log \left(\left[{H}^{+}\right]\right)$

$p {H}_{\text{sol}} = - \log \left(0.1\right) = \textcolor{g r e e n}{1}$