CH 115 Fall 2014
Worksheet 26
1. What is a solution composed of?
Solute – substance you are dissolving
Solvent – what you are dissolving the solute in
2. What is the formula for calculating molarity of a solution?
Molarity = moles of solute / liters of solution
3. Complete the following calculations.
a). What is the molarity of a solution made by dissolving 2.5 g of NaCl in enough
water to make 125 ml of solution?
2.5 g NaCl * (1 mol NaCl / 58.45 g NaCl) = 0.0427715997 moles NaCl
125 mL (1 liter / 1000 mL) = .125 L
Molarity = moles/L = 0.0427715997 moles/.125 L = 0.342 M
b). How many moles of salt are contained in 300 mL of a 0.40 M NaCl solution?
M = mol/L  mol = M * V
Moles = 0.40 M * .300 L = 0.12 moles salt
c). Battery acid is generally 3 M H2SO4. Roughly how many grams of H2SO4 are in
400 mL of this solution?
M = mol/L  mol = M*V
Mol = 3 M * .400 L = 1.2 moles H2SO4 * (98 g H2SO4 / 1 mol H2SO4) = 117.6 g
d). A chemist dissolves 98.4 g of FeSO4 in enough water to make 2.000 L of solution.
What is the molarity of the solution?
98.4 g FeSO4 * (1 mol FeSO4 / 151.85 g FeSO4) = 0.6480079025 mol FeSO4
M = 0.6480079025 mol / 2 L = 0.324 M
4. You have a stock solution that is 0.276 M. How much volume of this stock solution
do you need to add to a solution to make 100 mL of 0.128 M solution?
M1V1 = M2V2
V1 = M2V2 / M1 = (0.128 M)(100 mL) / 0.276 M = 46.38 mL
5. Write the balanced equation for the reaction of calcium carbonate with hydrochloric
acid. What mass of calcium carbonate is required to react with 25 mL of 0.75 M of
hydrochloric acid?
CaCO3 + 2 HCl  CO2 + H2O + CaCl2
.025 L * 0.75 M = 0.01875 mol HCl * (1 mol CaCO3 / 2 mol HCl) * (100.08 g
CaCO3 / 1 mol CaCO3) = 0.938 g CaCO3
6. What is a titration? What is the difference between the end point and the
equivalence point of a titration? What is special about the equivalence point?
A titration is a controlled experiment involving an acid-base neutralization
reaction. The end point of a titration is the point at which the indicator
CH 115 Fall 2014
Worksheet 26
changes colors. The equivalence point of a titration is the point at which the
moles of acid in the experiment equal the moles of base (M1V1 = M2V2).
7. A titration reveals that 11.6 mL of 3.0 M sulfuric acid are required to neutralize the
sodium hydroxide in 25.00 mL of NaOH solution. What is the molarity of the NaOH
solution?
H2SO4 + 2 NaOH  2 H2O + Na2SO4
0.0116 L * 3.0 M = 0.0348 moles H2SO4 * (2 moles NaOH / 1 mole H2SO4) =
0.0696 moles NaOH
M = moles/L = 0.0696 moles/ .025 L = 2.78 M