Question 3.14: REACTION STOICHIOMETRY IN SOLUTION Stomach acid, a dilute so...
REACTION STOICHIOMETRY IN SOLUTION
Stomach acid, a dilute solution of HCl in water, can be neutralized by reaction with sodium hydrogen carbonate, NaHCO_{3}, according to the equation
HCl(aq) + NaHCO_{3}(aq) → NaCl(aq) + H_{2}O(l) + CO_{2}(g)
How many milliliters of 0.125 M NaHCO_{3} solution are needed to neutralize 18.0 mL of 0.100 M HCl?
STRATEGY
Solving stoichiometry problems always requires finding the number of moles of one reactant, using the coefficients of the balanced equation to find the number of moles of the other reactant, and then finding the amount of the other reactant. The flow diagram summarizing the situation was shown in Figure 3.4.
We first have to find how many moles of HCl are in 18.0 mL of a 0.100 M solution by multiplying volume times molarity:
Moles of HCl = 18.0 \cancel{mL} × \frac{1 \cancel{L}}{1000 \cancel{mL}} × \frac{0.100 mol}{1 \cancel{L}} = 1.80 × 10^{-3} mol HCl
Next, check the coefficients of the balanced equation to find that 1 mol of HCl reacts with 1 mol of NaHCO_{3}, and then calculate how many milliliters of a 0.125 M NaHCO_{3} solution contains 1.80 × 10^{-3} mol:
1.80 × 10^{-3} \cancel{mol HCl} × \frac{1 \cancel{mol NaHCO_{3}}}{1 \cancel{mol HCl}} × \frac{1 L solution}{0.125 \cancel{mol NaHCO_{3}}} = 0.0144 L solution
Thus, 14.4 mL of the 0.125 M NaHCO_{3} solution is needed to neutralize 18.0 mL of the 0.100 M HCl solution.
BALLPARK CHECK
The balanced equation shows that HCl and NaHCO_{3} react in a molar ratio, and we are told that the concentrations of the two solutions are about the same. Thus, the volume of the NaHCO_{3} solution must be about the same as that of the HCl solution.
Neutralization of sodium hydrogen carbonate with acid leads to release of CO_{2} gas, visible in this fizzing solution.
