Write the balanced net ionic equations (c – j only) to illustrate the reaction(s
ID: 511145 • Letter: W
Question
Write the balanced net ionic equations (c – j only) to illustrate the reaction(s) occurring between the solute and water. Use an RICE box to calculate the theoretical pH value for each of these solutions. Question (c) has been started for you as an example. Write your answer for each solution in the box provided. Useful Equilibrium Constants
HC2H3O2(aq) Ka = 1.8x10-5
H3BO3(aq) Ka = 5.4x10-10
H2CO3(aq) Ka1 = 4.3x10-7, Ka2 = 5.6x10-11 H3PO4(aq)
Ka1 = 7.5x10-3 , Ka2 = 6.2x10-8, Ka3 = 4.2x10-13
NH3(aq) Kb = 1.8x10-5
a) Deionized water (DI-H2O) pH =
b) 0.10M HCl pH =
c) 0.10M HOAc (acetic acid)
d) 0.10 M NH4OAc (ammonium acetate) pH = 2
e) 0.10 M NH4Cl pH =
f) 0.10 M KH2PO4 pH =
g) 0.10 M H3BO3 pH = 3
h) 0.10 M Na2CO3 pH =
i) 0.10 M NaHCO3 pH =
j) 0.10 M Na2CO3 mixed with 0.10 M NaHCO3 pH =
Explanation / Answer
(a) Deionized water is water that has been passed through resins that remove electrically charged material from the water. If it has not come into contact with carbon dioxide, the pH of deionized water is 7.
(b)pH of 0.1M HCl
[H+] ion concentration = 10^-1
pH = -log[H+] = -log 10^-1 = 1
(c) pH of 0.1M acetic acid
[H+] concentration = 10^-1
pH. = 1
(d) pH of 0.1M ammonium acetate
CH3COONa is a basic salt :CH3COO- + Na+
CH3COO- + H2O <------> CH3COOH + OH-
Kb = Kw/Ka = 5.6 x 10^-10 = x^2/ 0.1-x
x = 7.5 x 10^-6 M = [OH-]
pOH = 5.1
pH = 14 - 5.1 = 8.9
(e) pH of 0.1 M
NH4+ <--> NH3 + H+
The equilibrium constant for weak acid is:
Ka = [NH3] [H+] / [NH4+] = 1×10^-14/1.8×10^-5
=5.69×10^-10
NH4+ -------- NH3+ H+
Initial x. 0. 0
At equilibrium
0.1-x. x. x
Ka. = [NH3] [H+]/[NH4+]
5.69×10^-10= x^2/(0.1-x)
x = 7.54×10^-6
= [NH3] = [H+] = 7.54×10^-6
pH=-log 7.54×10^6=6.123
(f) pH of 0.1M KH2PO4
H2PO4- + H2O ------ H3PO4 + OH-
Kb = Kw/Ka1 = 1.33x10^-12
[H+] = (Ka2×C) = ((6.2x10^-8)×(0.1)) = 7.9x10^-5
pH= -log 7.9×10^-5
pH = 4.1.
(g) pH of 0.1 M H3BO3
[H+] concentration =3× 10^-1
pH =- -log 3×10^-1=0.5229
(h) pH of 0.1M Na2CO3
Na2CO3 --> 2Na+ + (CO3)-2
Then the CO3-2 anion reacts with water:
CO3-2 + H2O <==> HCO3-+ OH-
Kb = [HCO3-][OH-]/[CO3-2][H2O]
2×10^(-4) = x^2/[0.1M][1]
x = 4.47 ×10^-3
pOH = -log[OH-]=2.35
pH = 14 -pOH= 11.65
(i) pH of 0.1M NaHCO3
Bicarbonate ions are amphiprotic having both acidic and basic reactions:
HCO3- + H2O <----> H2CO3 + OH- (base)
HCO3- <----> CO3^2- + H+ (acid)
both reactions will occur, and both must be included in calculating the [H+] concentration.
[H+] = [Ka1×Ka2] = 4.65×10^-9
pH =-log 4.65×10^-9= 8.33
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