Question 4.10: Neutralization Reactions I What volume of a 0.100 M HCl solu...

Where are we going?
We are asked to determine the volume of a solution of HCl required to neutralize a given volume of a solution of NaOH.

What do we know?
We know the concentration and volume of the NaOH solution and the concentration of the HCl solution.

How do we get there?
The HCl and NaOH 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 present in the mixed solutions before any reaction occurs are

H^{+},  Cl^{-},                  Na ^{+}, OH ^{-} ,             and    H_{2} O

From HCl(aq)            From NaOH(aq)

Which reaction will occur? The two possibilities are

Na ^{+}(aq) +  Cl^{-}(aq) → NaCl(s)

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

Since NaCl is soluble, the first reaction does not take place (Na ^{+} and Cl^{-} are spectator ions). However, as we have seen before, the reaction of H^{+} and OH ^{-} ions to form H_{2} O does occur.
The balanced net ionic equation for the reaction is

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

Next, we calculate the number of moles of OH ^{-} ions in the 25.0-mL sample of 0.350 M NaOH:

25.0   mL   NaOH  ×\frac{1   L }{1000   mL} × \frac{0.350   mol  OH ^{-}}{ L  NaOH}    =8.75 × 10 ^{-3}  mol  OH ^{-}

This problem requires the addition of just enough H^{+} ions to react exactly with the OH ^{-} ions present. Thus we need not be concerned with determining a limiting reactant.
Since H^{+} and OH ^{-} ions react in a 1;1 ratio, 8.75 × 10 ^{-3} mole of H^{+} ions is required to neutralize the OH ^{-}  ions present.
The volume (V) of 0.100 M HCl required to furnish this amount of H^{+} ions can be calculated as follows:

V × \frac{0.100   mol  H^{+} }{L} = 8.75 × 10 ^{-3}   mol   H^{+}

Solving for V gives

V =  8.75  ×  10 ^{-2}L

Thus 8.75  × 10 ^{-2}  L (87.5 mL) of 0.100 M HCl is required to neutralize 25.0 mL of 0.350 M NaOH.

Is the answer reasonable?
The mole ratio between the reactants is 1;1. The NaOH is more concentrated than the HCl, so it makes sense that the volume of HCl is larger than the volume of NaOH.