# Question 2b095

May 30, 2015

You'd need 30 mL of hydrochloric acid solution to get the pH to that value.

So, you know that you're dealing with a potassium hydroxide solution. Since potassium hydroxide is strong base, it dissociates completely in aqueous solution to give potassium cations, ${K}^{+}$, and hydroxide anions, $O {H}^{-}$.

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

The $1 : 1$ mole ratio that exists between $K O H$ and $O {H}^{-}$ implies that the concentration of the hydroxide ions will be equal to that of the potassium hydroxide

$\left[O {H}^{-}\right] = \left[K O H\right] = \text{0.05 M}$

Use the volume of the solution to determine how many moles of hydroxide ions are present

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

${n}_{O {H}^{-}} = \text{0.05 M" * 20 * 10^(-3)"L" = "0.001 moles}$ $O {H}^{-}$

This means that, in order to get a neutral solution, you need to add 0.001 moles of hydrochloric acid. This much acid will completely neutralize the initial solution and get it to a pH of 7.

Figure out how much volume of hydrochloric acid solution is needed to neutralize the potassium hydroxide solution

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

${V}_{\text{neutralization" = "0.001 moles"/"0.05 M" = "0.02 L" = "20 mL}}$

However, you want the final solution to be acidic, which implies that you need an excess of ${H}_{3} {O}^{+}$.

Use the final solution's pH to determine what the concentration of the hydronium ions must be in order for the solution to have pH of 2

pH_"sol" = -log([H_3O^(+)]) => [H_3O^(+)] = 10^(-pH_"sol")

$\left[{H}_{3} {O}^{+}\right] = {10}^{- 2} = \text{0.01 M}$

Now you need to determine how much $H C l$ solution you need to add to add to get that concentration of hydronium ions.

Think of it like this. After neutralization, the number of moles of hydrochloric acid present in what you add must be equal to the number of moles of hydrochloric acid that you're left with.

${n}_{\text{HCl added" = n_"HCl final}}$

In other words, you need to dilute a sample of $H C l$ of known concentration to a solution of known volume and concentration

${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, where

${C}_{1}$ - the molarity of the $H C l$ solution;
${V}_{1}$ - the volume you need to add after the initial solution was neutralized;
${C}_{2}$ - the concentration of the final solution;
${V}_{2}$ - the total volume of the final solution.

But ${V}_{2}$ is actually what you already have plus what you need to add

${V}_{2} = {V}_{1} + 20 + 20 = {V}_{1} + \text{40 mL}$

This means that you get

${V}_{1} = {C}_{2} / {C}_{1} \cdot \left({V}_{1} + \text{40 mL") = (0.01cancel("M"))/(0.05cancel("M")) * (V_1 + "40 mL}\right)$

${V}_{1} = 0.2 \cdot {V}_{1} + \text{8 mL" => V_1 = "8 mL"/0.8 = "10 mL}$

Therefore, the total volume of $H C l$ solution you need to add will be

V_"HCl total" = V_"neutralization" + V_1 = 20 + 10 = color(green)("30 mL")#