Question 6.17: 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
\mathrm{HCl}(a q)+\mathrm{NaHCO}_3(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_2 \mathrm{O}(l)+\mathrm{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 chart summarizing the situation was shown in Figure 6.7.
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:
\text{Moles of HCL} = 18.0 \ \cancel{mL} \times \frac{1 \ \cancel{L}}{1000 \ \cancel{mL}} \times \frac{0.100 \ mol}{1 \ \cancel{L}} = 1.80 \times 10^{-3} \ mol \ HCLNext, 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 \times 10^{-3} mol:
1.80 \times 10^{-3} \ \cancel{mol \ HCl} \times \frac{1 \ \cancel{mol \ NaHCO_3}}{1 \ \cancel{mol \ HCl}} \times \frac{1 \ L \ solution}{0.125 \ \cancel{mol \ NaHCO_3}} = 0.0144 \ L \ solutionThus, 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 1:1 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.