# Question 3aabb

May 28, 2015

The solution will be basic.

Start with the balanced chemical equation for this neutralization reaction

${H}_{2} S {O}_{4 \left(a q\right)} + C a {\left(O H\right)}_{2 \left(a q\right)} \to C a S {O}_{\textrm{4 \left(s\right]}} + 2 {H}_{2} O$

Notice that you have a $1 : 1$ mole ratio between sulfuric acid and calcium hydroxide. This tells you that, in order for a complete neutralization to take place, you need equal amounts of moles of the two compounds.

If you have more moles of acid than of base, the solution will have an excess of hydronium ions, ${H}_{3} {O}^{+}$, and will therefore be acidic.

If you have more moles of base, the solution will hve an excess of hydroxide ions, OH""^(-), and will therefore be basic.

Use the molarities and volumes of the two solutions to determine how many moles of each you have

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

${n}_{{H}_{2} S {O}_{4}} = \text{0.25 M" * 45 * 10^(-3)"L" = "0.01125 moles}$ ${H}_{2} S {O}_{4}$

and

${n}_{C a {\left(O H\right)}_{2}} = \text{0.48 M" * 25 * 10^(-3)"L" = "0.01200 moles}$ $C a {\left(O H\right)}_{2}$

You have more moles of calcium hydroxide present, which means that the sulfuric acid will be completely consumed by the reaction and you'll only be left with

${n}_{{H}_{2} S {O}_{4}} = 0$

${n}_{C a {\left(O H\right)}_{2}} = 0.01200 - 0.01125 = \text{0.00075 moles}$ $C a {\left(O H\right)}_{2}$

Therefore, the solution will be basic.

If you need to calculate the pH as well, use the dissociation of calcium hydroxide to get the number of moles of hydroxide ions present

$C a {\left(O H\right)}_{2} \to C {a}^{2 +} + \textcolor{red}{2} O {H}^{-}$

You'll get twice as many moles of hydroxide ions than you have of the remaining calcium hydroxide. Use the total volume of the solution to determine the molarity of the $O {H}^{-}$ ions.

From there, determine pOH and pH.

May 29, 2015

I don't want the initial answer to become very long, so I'll just show you how to get the solution's pH here.

So, you know that your reaction left you with excess calcium hydroxide, which is a strong base. In your case, calcium hydroixde will dissociate completely to produce calcium cations, $C {a}^{2 +}$, and hydroixde anions, $O {H}^{-}$, according to the following equation

$C a {\left(O H\right)}_{2 \left(a q\right)} \to C {a}_{\textrm{\left(a q\right]}}^{2 +} + \textcolor{red}{2} O {H}_{\left(a q\right)}^{-}$

Notice that you get $\textcolor{red}{2}$ moles of hydroxide ions for every mole of calcium hydroxide that dissociates. SInce you were left with 0.00075 moles of $C a {\left(O H\right)}_{2}$ in solution, the number of moles of $O {H}^{-}$ will be

0.00075cancel("moles"Ca(OH)_2) * (color(red)(2)" moles "OH^(-))/(1cancel("mole"Ca(OH)_2)) = "0.00150 moles"#

To get the molarity of the hydroxide ions, use the total volume of the solution

${V}_{\text{total}} = {V}_{{H}_{2} S {O}_{4}} + {V}_{C a {\left(O H\right)}_{2}}$

${V}_{\text{total" = 45 + 25 = "70. mL}}$

This will get you

$\left[O {H}^{-}\right] = \text{0.00150 moles"/(70. * 10^(-3)"L") = "0.02143 M}$

SIDE NOTE Don't forget to convert mL to L when calculating molarity!

Since you know the concentration of the hydroxide ions, you can determine the solution's pOH by

$p O H = - \log \left(\left[O {H}^{-}\right]\right)$

$p O H = - \log \left(0.02143\right) = 1.67$

THis means that the solution's pH will be

$p {H}_{\text{sol}} = 14 - p O H = 14 - 1.67 = \textcolor{g r e e n}{12.33}$