In-Class Group Activity #7 (Chem 101, Summer 2006)
Name ______KEY______
1. A Brønstead-Lowry ______acid______donates H+ and a Brønstead-Lowry
______base______accepts H+.
2. Give the conjugate base for each of the following Brønstead-Lowry acids:
a) HIb) H2SO4c) HCO3-d) H2O
I-HSO4-CO32-OH-
3. Give the conjugate acid for each of the following Brønstead-Lowry bases:
a) H2PO4-b) NH3c) OH-d) H2O
H3PO4NH4+H2OH3O+
4. For the following reaction: NH3 + H2O NH4+ + OH-
a) Identify which reactant is the acid and which is the base.
Acid = H2OBase = NH3
b) Identify the conjugate acid and conjugate base.
Conjugate acid = NH4+Conjugate base = OH-
c) Write the Kb expression for the reaction.
Kb = [NH4+][OH-]/[NH3]
d) Does the equilibrium favor reactants or products (Kb of NH3 = 1.8 x 10-5) ?
Reactants
5. Since H2O can act as either a weak acid or a weak base, it can react with itself.
a) Write a balanced chemical equation for the reaction.
2H2O H3O+ + OH-
b) Does the equilibrium favor reactants or products?Reactants
c) What is Kw (write the expression and give the value)?
Kw = [H3O+][OH-] = 1.0 x 10-14
d) What is the concentration of OH- in pure water?1.0 x 10-7 M
6. Consider the following two reactions:
A. HCl + H2O H3O+ + Cl-
B. CH3CO2H + H2O CH3CO2- + H3O+
a) Which reaction involves a strong acid, and which one a weak acid?
Reaction A involves a strong acid and reaction B involves a weak acid.
b) If 1.0 g of HCl is dissolved in H2O to a total volume of 250 mL, what is the pH of the solution?
Since HCl is a strong acid it completely ionizes in water according to reaction A.
Plan: g HCl mol HCl mol H3O+ M H3O+ pH
1.0 g HCl x (1 mol/36.46 g) x (1 mol H3O+/1 mol HCl) = 0.0274 mol H3O+
[H3O+] = 0.0274 mol H3O+/0.250 L = 0.11 M H3O+
pH = -log[H3O+] = - log(0.11) = 0.96
c) Write the Ka expression for reaction B.
Ka = [CH3CO2-][H3O+]/[CH3CO2H]
d) What would you add to reaction B to make a buffer?
To make a buffer, add a salt containing the conjugate base, such as NaCH3CO2.
e) If the Ka for acetic acid (CH3CO2H) is 1.8 x 10-5, what is the pH of a buffer solution prepared from 1.2 M CH3CO2H and 0.90 M CH3CO2- ?
[H3O+] = Ka x [CH3CO2H]/[CH3CO2-] = (1.8 x 10-5)(1.2/0.90) = 2.4 x 10-5 M
pH = -log[H3O+] = -log(2.4 x 10-5) = 4.62
7. What is the pH of a solution that has an OH- concentration of 1.0 x 10-3 M?
Kw = [H3O+][OH-] = 1.0 x 10-14
[H3O+] = 1.0 x 10-14/[OH-] = 1.0 x 10-14/1.0 x 10-3 = 1.0 x 10-11
pH = -log[H3O+] = -log(1.0 x 10-11) = 11.00
8. Complete and balance the following reactions:
a) Ca + 2HNO3 Ca(NO3)2 + H2
b) KHCO3 + HCl KCl + CO2 + H2O
c) CH3CO2H + NaOH NaCH3CO2 + H2O
9. State whether each of the following salts would make a neutral, basic or acidic solution when dissolved in water. If the solution would be basic or acidic, write the equation for the reaction that causes the pH change:
a) Na2SO4 (comes from NaOH + HSO4-)
Basic: SO42- + H2O HSO4- + OH-
(NaOH is a strong base and HSO4- is a weak acid. *Sorry for the mistake in the original question.)
b) NH4Br (comes from NH3 + HBr)
Acidic: NH4+ + H2O NH3 + H3O+
(NH3 is a weak base and HBr is a strong acid.)
c) KCH3CO2 (comes from KOH + CH3CO2H)
Basic: CH3CO2- + H2O CH3CO2H + OH-
(KOH is a strong base and CH3CO2H is a weak acid.)
10. A phosphate buffer is made by dissolving H3PO4 and NaH2PO4 in water.
a) Write the equation that shows how the buffer would neutralize a small amount of added HCl (remember HCl + H2O H3O+ + Cl-):
H2PO4- + H3O+ H3PO4 + H2O
b) Write the equation that shows how the buffer would neutralize a small amount of added NaOH (remember NaOH Na+ + OH-):
H3PO4 + OH- H2PO4- + H2O