Acid – Base Test Review
1. Be able to define solvent, solute, solution, colloid, suspension, hydrogen bond, solubility, molarity, surfactant,
surface tension, electrolyte, Tyndall effect, saturated solution, supersaturated solution, unsaturated solution,
miscible, immiscible, dilute solution, concentrated solution, Arrhenius acid/base, Bronsted-Lowry acid/base, Lewis
acid/base, conjugate acid/base.
2.
a) What is the solubility of KCl at 10 degrees C?
30 g/100 g H2O
b) Which compound shown has decreasing solubility as
temperature increases?
cerium (III) sulfate
c) Which substance shows the least change in
solubility?
NaCl
d) Suppose you added 10 grams of KCl to a saturated
solution at 40 degrees C and then warmed it up to 90
degrees C. What would happen?
The added KCl would dissolve as the soln. is heated.
3. Explain why solid water (ice) is less dense than liquid water. Is this a normal phenomenon?
As water freezes, the hydrogen bonds between water molecules cause them to arrange in a hexagonal
pattern with empty space in the middle. The empty space within this rigid framework causes the density to drop. It
is not common for the solid form of a substance to be less dense than the liquid form.
4. An aqueous system is not uniform throughout and has particles larger than 1 nm. What type is it? colloid
5. Which definition would you use to classify the following as bases?
a. KOH  K+ + OH- Arrhenius because the base fully dissociates into its ions
b. PCl3 Lewis because the compound has a pair of nonbonding electrons that it can donate. (HINT: Draw the
Lewis structure.)
6. What is the molarity of 25 mL HCl containing 0.60 mol? mol/L = 0.60mol/0.025L = 24M
7. What mass of HBr would you need to make 100 mL of 1.3 M HBr? mol = MxL = 1.3M x 0.100L = 0.13 mol
0.13 mol HBr x 80.91 g HBr/ 1 mol HBr = 10.52 g HBr
8. To what volume would you need to dilute 40 mL of 0.50 M KOH to make a 0.1 M solution? M1V1 = M2V2,
V2 = M1V1/M2 = 0.50 M x 40 mL / 0.1 M = 200 mL KOH
so
9. What is the pH of the following solutions?
a. 0.95 M HNO3 pH = -log(0.95) = 0.02
b. 0.02 M NaOH pOH = -log(0.02) = 1.70, so 14 – 1.70 = pH = 12.30
10. What is the [OH-] of a 1.4 M solution of HCl? Kw = [H+][OH-], so Kw/[H+] = [OH-] = 1.0 x 10-14 / 1.4 = 7.14 x 10-15M
11. Complete the following neutralization equations and identify the conjugate acids and bases:
a. 2NaOH + H2SO4  Na2SO4 + 2HOH
A
B
CB
CA
b. HI + KOH  KI + HOH
A
B
CB
CA
12. If 38 mL of 0.5 M NaOH is used to titrate 10.00 mL of HCl, what is the molarity of the HCl?
Ma = ?, Va = 10.00 mL, Mb = 0.5 M, Vb = 38 mL
HCl + NaOH  NaCl + HOH
1) Find moles of base: mol = M x L = 0.5 M x 0.038L = 0.019 mol NaOH
2) Convert to moles of acid: 0.019 mol NaOH x 1 mol HCl / 1 mol NaOH = 0.019 mol HCl
3) Calculate molarity of acid:
M = mol/L = 0.019 mol HCL / 0.01000 L HCl = 1.9 M HCl