MathConnections
CH 3 CHCH 2 CH3CH2CH2CH2
4.5 Solution Stoichiometry
In the stoichiometry problems we have discussed so far, the numbers of moles in- volved have been obtained by using the ratio established by the molar mass of the substances. Now we can expand the range of problems that we can consider by us- ing Equation 3.2 to determine the number of moles of substances when volumes are measured rather than masses. The heart of the stoichiometry problem, however, remains the same. The balanced chemical equation still provides the critical ratios among numbers of moles of various species in the reaction. Again, a fl ow diagram for solving a stoichiometry problem shows the manipulations we need to consider (Figure 4.5).
Volume and concentration
(given)
Desired answer (molarity or
volume) n = M β«» V
Balanced chemical equation RATIO Moles
in
n = M β«» V Moles
out
Figure 4.5 β This fl ow diagram shows the important steps in a typical solution stoichiometry calculation.
The only complication added by considering reactions in solution is the need to relate three variables, using Equation 3.2, rather than two variables using the molar mass ratio. Example Problem 4.9 shows how to approach these solution stoichiometry problems.
E X A M P L E P RO B L E M 4 . 9
As we mentioned in Example Problem 3.10, the fuel hydrazine can be produced by the reaction of solutions of sodium hypochlorite and ammonia. The relevant chemical equation is
NaClO(aq) + 2 NH3(aq) : N2H4(aq) + NaCl(aq) + H2O(π΅)
If 750.0 mL of 0.806 M NaClO is mixed with excess ammonia, how many moles of hydrazine can be formed? If the fi nal volume of the resulting solution is 1.25 L, what will be the molarity of hydrazine?
Strategy We are asked for the expected amount of a product, so this is a reaction stoichiometry problem. Because NH3 is said to be in excess, we know that NaClO will be the limiting reactant. So we will use the given volume and concentration to fi nd the number of moles of NaClO reacting. Then use the mole ratio from the balanced equation to fi nd the number of moles of N2H4 that can be formed. Finally, we can use that number of moles and the given fi nal volume to obtain the molarity.
Solution Convert 750.0 mL to 0.7500 L and determine the number of moles of NaClO reacting:
nNaClO= M Γ V = 0.806 mol/L Γ 0.7500 L = 0.605 mol NaClO
Use that along with the 1:1 mole ratio from the balanced equation to fi nd the corresponding number of moles of hydrazine:
0.605 mol NaClO Γ 1 mol N2H4
βββββββ
1 mol NaClO = 0.605 mol N2H4
To answer the second part of the problem, we just need to use this number of moles along with the given fi nal volume to fi nd the molarity.
M = n ββ
V = 0.605 mol N2H2
ββββββββ
1.25 L = 0.484 M N2H4
Discussion Notice that the stoichiometric calculation is done using the mole ratio from the balanced equation, just as in all of our earlier examples. Only the conversions to and from numbers of moles have changed because this reaction is in solution.
Check Your Understanding What volume (in milliliters) of 0.150 M HCl is required to react completely with 0.503 g of dry Na2CO3?
A common laboratory technique, called titration, requires understanding solution stoichiometry. A solution-phase reaction is carried out under controlled conditions so that the amount of one reactant can be determined with high precision. A care- fully measured quantity of one reactant is placed in a beaker or fl ask. A dye called an indicator can be added to the solution. The second reactant is added in a controlled fashion, typically by using a burette (Figure 4.6). When the reaction is complete, the indicator changes color. When the indicator fi rst changes color, we have a stoichio- metric mixture of reactants. We know the number of moles of the fi rst reactant (or the molarity and volume) and the volume of the second reactant used. So as long as we know the balanced equation for the reaction, we can fi nd the unknown concentration of the second reactant.
Indicators are also examples of weak acids or bases, as defi ned in Chapter 3.
Indicators are also examples of weak acids or bases, as defi ned in Chapter 3.
E X A M P L E P RO B L E M 4 .10
Many common titrations involve the reaction of an acid with a base. If 24.75 mL of 0.503 M NaOH solution is used to titrate a 15.00-mL sample of sulfuric acid, H2SO4, what is the concentration of the acid?
Strategy A titration problem is an applied stoichiometry problem, so we will need a balanced chemical equation. We know the molarity and volume for the NaOH solu- tion, so we can fi nd the number of moles reacting. The mole ratio from the balanced equation lets us calculate moles of H2SO4 from moles of NaOH. Because we know the volume of the original H2SO4 solution, we can fi nd its molarity.
Solution The acidβbase reaction will produce sodium sulfate and water. The bal- anced equation is
H2SO4(aq) + 2 NaOH(aq) : Na2SO4(aq) + 2 H2O(π΅)
Determine the number of moles of NaOH from the given molarity and volume:
0.02475 L solution Γ 0.503 mol NaOH
βββββββββ
1 L solution = 0.0124 mol NaOH Use the balanced chemical equation to determine the number of moles of H2SO4:
0.0124 mol NaOH Γ 1 mol H2SO4
βββββββ
2 mol NaOH = 6.22 Γ 10β3 mol H2SO4 Determine the concentration of H2SO4:
M = 6.22 Γ 10β3 mol H2SO4
βββββββββββ
0.01500 L solution = 0.415 M H2SO4
Analyze Your Answer The molarity we found for sulfuric acid is smaller than that we had for sodium hydroxide. Does this make sense? The point to remember here is the stoichiometry that provides the 2:1 mole ratio. Even though the volume of NaOH used is greater than the volume of the acid, it is not twice the volume, so our H2SO4 molarity should be smaller.
Discussion This is the type of problem that students get wrong if they use the di- lution formula (Eq. 3.3) by mistake. The dilution formula does not include the mole ratio from a balanced chemical equation, and in this case, the 2:1 ratio would result in an error of a factor of 2 if you tried to use Eq. 3.3.
Check Your Understanding Analysis of a solution of NaOH showed that 42.67 mL of 0.485 M HNO3 was needed to titrate a 25.00-mL sample of NaOH.
What is the concentration of the NaOH solution?
Figure 4.6 β The photo sequence shows some of the steps in a typical titration. One solution is poured into a burette in the left-hand panel.
The second solution is placed into an Erlenmeyer fl ask, and the burette is positioned above the fl ask in the second panel. The valve on the burette allows controlled addition of solution, and the shape of the Erlenmeyer fl ask permits easy mixing by swirling. In the fi nal panel, the solution in the fl ask turns pink when the endpoint is reached.
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