Question 4.11: Neutralization Reactions II In a certain experiment, 28.0 mL...

Where are we going?
We are asked to determine the amount of water formed and the amount of excess reactant when an acid and base react.

What do we know?
We know the volumes and concentrations of the reactants.

How do we get there?
The acid and base react with each other, so we need to use a balanced equation to determine the mole ratio between the two reactants. Since we have volumes and molarities of both, we can determine the moles of each reactant. To do this, we must think about the components in the reactant solutions.
The species available for reaction are

H^{+},  NO_{3} ^{-}  ,              K^{+}, OH^{-} ,                 and H  _{2} O
From HNO_{3}              From KOH
solution                        solution

Since KNO_{3} is soluble, K^{+} and NO_{3} ^{-} are spectator ions, so the net ionic equation is

H^{+}(aq)  +  OH^{-}(aq)  →  H  _{2} O (l)

We next compute the amounts of H^{+} and OH^{-} ions present.

28.0   mL   HNO_{3}  ×  \frac{1   L }{1000   mL} ×  \frac{0.250   mol  H^{+}}{L}  = 7.00 ×  10 ^{-3}  mol    H^{+}

  53.0   mL   KOH   ×  \frac{1   L }{1000   mL}  ×  \frac{0.320      mol  OH^{-}}{L}  =  1.70  ×  10 ^{-2}    mol   OH^{-}

Since H^{+} and OH^{-} react in a 1;1 ratio, the limiting reactant is H^{+}. Thus 7.00 × 10 ^{-3} mole of H^{+} ions will react with 7.00 × 10 ^{-3} mole of OH^{-} ions to form 7.00 × 10 ^{-3} mole of H  _{2} O.
The amount of OH^{-} ions in excess is obtained from the following difference:

Original amount - amount consumed = amount in excess

1.70  ×  10 ^{-2}    mol   OH^{-}  –  7.00 × 10 ^{-3} mol  OH^{-} = 1.00  ×  10 ^{-2}    mol   OH^{-}

The volume of the combined solution is the sum of the individual volumes:

Original volume of HNO_{3}  +  original volume of KOH = total volume

28.0 mL + 53.0 mL = 81.0 mL = 8.10  ×  10 ^{-2} L

Thus the molarity of OH^{-} ions in excess is

M = \frac{ mol  OH^{-}}{L   solution}

= \frac{ 1.00  ×  10 ^{-2}  mol  OH^{-}}{8.10  ×  10 ^{-2}  L}

= 0.123 M  OH^{-}

Is the answer reasonable?
The mole ratio between the reactants is 1;1. The KOH is more concentrated than the HNO_{3} , and a greater volume of KOH is used. It makes sense that the OH^{-} is in excess.