Lab Mini-quiz 4 Solutions

Chem 5
Fall 2014
Mini quiz 4 – Solutions
Mon Formic acid, HCOOH, is a simple weak acid. It dissociates into H+ and the formate anion,
HCOO– as shown below. (Note that the acid has two hydrogens, but only one of them is
involved in the acid/base chemistry of formic acid. You’ll learn in Chem 6 why this is so.)
→ H+(aq) + HCOO–(aq)
HCOOH(aq) ←
Ka = 1.77 × 10–4
(a) Calculate the pH of a solution made by dissolving 0.20 mol HCOOH and 0.15 mol
NaHCOO (which is sodium formate, a solid compound that dissociates completely in
water) in 500 mL of pure water.
Because Ka is quite small, we can assume that most of the formic acid we start with, 0.20
mol, stays intact with very little dissociation. This expectation is strengthened when we
realize that the formate ion we’re adding from NaHCOO, 0.15 mol, drives the acid
dissociation reaction in a direction that further inhibits HCOOH dissociation without
appreciably lowering the formate ion amount. Note as well that both these species,
HCOOH and NaHCOO, are in the same 500 mL volume so that we can use the ratio of
their amounts in mols in place of the ratio of their concentrations in our equilibrium
constant expression:
Ka =
[H+][HCOO–] [H+] 0.15 mol/0.500 L [H+] 0.15 mol
= 1.77 × 10–4
0.20 mol/0.500 L
0.20 mol
We solve for [H+] and find [H+] = 2.36 × 10–4 M or a pH = –log10(2.36 × 10–4) = 3.63.
(b) Write the predominant chemical reaction that occurs when 0.20 mol NaOH(s) is added
to this solution.
NaOH(s) dissolves to produce Na+(aq) and the strong base OH–(aq). This base reacts
with the undissociated formic acid, (and note that we have added as many moles of OH– as
we had of formic acid originally):
→ HCOO–(aq) + H2O(l)
HCOOH(aq) + OH–(aq) ←
(c) Which phrase below best characterizes the solution that results after this NaOH
addition? (Circle one.)
strongly acidic
weakly acidic
neutral (pH = 7)
weakly basic
strongly basic
In part (b), we added as many moles of NaOH as we had moles of HCOOH, turning them
all into HCOO– ions. It is as if we had added only NaHCOO to pure water. A solution of
predominantly HCOO– ions will be weakly basic due to hydrolysis of the weak base
Acetic acid, CH3COOH, is a simple weak acid we’ve probably all ingested in vinegar,
which is mostly acetic acid. It dissociates into H+ and the acetate anion, CH3COO– as
shown below. (Note that the acid has four hydrogens, but only one of them is involved in
the acid/base chemistry of acetic acid. You’ll learn in Chem 6 why this is so.)
→ H+(aq) + CH3COO–(aq)
CH3COOH(aq) ←
Ka = 1.76 × 10–5
Chem 5
Fall 2014
Mini quiz 4 – Solutions
(a) Calculate the pH of a solution made by dissolving 0.30 mol CH3COOH in 750 mL of
pure water.
This solution has an initial acetic acid concentration given by [CH3COOH]0 = (0.30
mol)/(0.75 L) = 0.40 M. The small value of Ka tells us that at equilibrium, not much of
this molecular acetic acid will have dissociated. But if we say x mole per liter does
dissociate, the equilibrium concentration of H+ and CH3COO– will both be x. Moreover, it
will be a good approximation to say the equilibrium acetic acid concentration, which is
strictly [CH3COOH]0 – x, will remain unchanged from its initial value. We have
= x = 1.76 × 10–5
[CH3COOH]0 0.40
1.76 × 10 0.40 = 2.65 × 10–3 M or pH = –log10 2.65 × 10–3 = 2.58
Ka =
(b) Next, 0.15 mol of sodium acetate, NaCH3COO, which completely dissociates into
Na+ and CH3COO– ions, is added to this solution. Now what’s the pH?
The Na+ ions do nothing, and the acetate ions influence the acetic acid dissociation reaction
(by inhibiting dissociation). We find [CH3COO–]0 = (0.15 mol)/(0.75 L) = 0.20 M. Our
acid dissociation equilibrium constant expression now looks like this:
[H+][CH3COO–] x [CH3COO–]0 x 0.20
= 1.76 × 10–5
1.76 × 10–5 0.40
= 3.52 × 10–5 M or pH = –log10 3.52 × 10–5 = 4.45
Ka =
We took a solution at pH 2.58, added a significant amount of a weak base (the acetate ion),
and sure enough, the pH rose; the solution became more basic.
Wed Hypochlorous acid, HClO, is a simple and very weak acid used to disinfect water supplies
and swimming pools. It dissociates into H+ and the hypochlorate anion, ClO– as shown
→ H+(aq) + ClO–(aq)
HClO(aq) ←
Ka = 3.0 × 10–8
(a) A common source for hypochlorate ion is the compound calcium hypochlorate,
Ca(ClO)2, a solid that is very soluble and that dissociates completely in water. Calculate
the pH of a solution made by dissolving 0.0122 mol Ca(ClO)2 (which is about 1.75 g) in
800 mL of water.
First, we note that 0.0122 mol Ca(ClO)2 is 2 × 0.0122 mol = 0.0244 mol ClO– anions.
Thus, [ClO–]0 = (0.0244 mol)/(0.800 L) = 0.0306 M. Next, we recognize what happens
chemically to these hypochlorate ions: they are hydrolyzed:
→ HClO(aq) + OH–(aq)
ClO–(aq) + H2O(l) ←
Kb = K w =
= 3.33 × 10–7
3.0 × 10
Chem 5
Fall 2014
Mini quiz 4 – Solutions
This small value tells us that the amount of hydrolysis will be small, making the
approximation that [ClO–] = [ClO–]0 a good one. Moreover, we expect [HClO] = [OH–] to
be a similarly good approximation. We write
x = [OH–] =
[HClO] [OH–]
3.33 × 10–7 0.0306 = 1.00 × 10–4 M
Kb = 3.33 × 10–7 =
This value for the hydroxide concentration, [OH–], gives us the pOH directly: pOH =
–log10(1.00 × 10–4) = 4.00 and then pH = 14 – pOH = 10.0. This is a decidedly basic
(b) If a strong acid such as HCl is added to this solution, Le Chatelier tells us that the
HClO dissociation reaction above would be driven back toward the HClO reactant. What
would be the pH if half as much of the ClO– is turned into HClO when H+ is added?
If we start with [ClO–] = 0.0306 M and [HClO] = 1.00 × 10–4 M = 0.0001 M, we see that
the amount of HClO is very small compared to the amount of ClO–. Thus, turning half the
ClO– into HClO will make the concentrations of these two species effectively equal to each
other: [HClO] = [ClO–] = (0.0306 M)/2 = 0.0153 M. We write
Ka =
[H+] = Ka = 3.0 × 10–8
or pH = PKa = –log10(3.0 × 10–8) = 7.52. Note that this is still a basic solution (pH > 7),
but not as basic as it was before we added the strong acid.