Worksheet # 5
Acid and Base pH Calculations and Stoichiometry
For each weak bases calculate the [OH-], [H+], pOH and pH. Remember that you need to calculate Kb first.
1.
0.20 M
CN- (aq) + H2O (l)
→
HCN(aq)
+ OH –(aq)
2
After setting up an ICE chart you get Kb= X / (0.20 – X) where the X is insignificant and ignored as it
passes the test of i/Kb >500 or 0.20/ 2.04 X 10-5 = 9804 which is bigger than 500 so X is ignored in (0.200 –X) .
1.6 X 10-5 is Kb from
(1.00 x 10-14/ Ka of HCN (6.2 x 10-10))
1.61 X 10-5 = X2 / 0.20
3.22 x 10-6 = X2
0.00179 = X or concentration of Base OHpOH = 2.75 and pH = 11.25 [H3O+] = 4.95 x 10-12
2.
0.010 M NaHS (the basic ion is HS-) HS- (aq) + H2O (l)
→
H2S (aq) + OH –(aq)
test of i/Kb >500 or 0.010/ 1.0 X 10-7 = 100,000 and is bigger than 500, X is ignored in (0.200 –X) .
1.1 X 10-7 is Kb from (1.00 x 10-14/ Ka of H2S (1.1 x 10-7))
1.1 X 10-7 = X2 / 0.010
9.1 X 10-8 = X2
3.01 X 10-5 = X or concentration of Base OHpOH = 4.52 and pH = 9.48 [H3O+] = 3.3 x 10-10
3.
0.067 M KCH3COO
CH3COO- (aq) + H2O (l)
→
CH3COOH (aq)
+ OH –(aq)
-10
test of i/Kb >500 or 0.067/ 5.5 X 10 = answer is bigger than 500, X is ignored in (0.200 –X) .
5.5 X 10-10 is Kb from (1.00 x 10-14/ Ka of CH3COOH 1.8 x 10-5)
5.5 X 10-10 = X2 / 0.067
3.68 X 10-11 = X2
6.1 X 10-6 = X or concentration of Base OHpOH = 5.21 and pH = 8.78
[H3O+] = 1.65 x 10-9
4.
0.40 M KHCO3
HCO3- (aq) + H2O (l)
→
H2CO3(aq) + OH –(aq)
test of i/Kb >500 or 0.40/ 2.1 X 10-4 = 1920 is bigger than 500, X is ignored in (0.200 –X) .
Kb = (1.00 x 10-14/ Ka 4.4 x 10-7)
2.3 X 10-8 = X2 / 0.40
9.09 X 10-9 = X2
9.5 X 10-5 = X or concentration of Base OHpOH = 4.02 and pH = 9.97
[H3O+] = 1.04 x 10-10
5.
0.60 M NH3
NH3 (aq) + H2O (l)
→
NH4(aq)+ + OH –(aq)
test of i/Kb >500 or 0.60/ 1.8 X 10-5 =33,333 is bigger than 500, X is ignored in (0.200 –X) .
1.8 X 10-5 is Kb from (1.00 x 10-14/ Ka (aq) 5.6 x 10-10)
1.8 X 10-5 = X2 / 0.60
1.08 X 10-5 = X2
0.00329 = X or concentration of Base OHpOH = 2.48 and pH = 11.52
[H3O+] = 3.04 x 10-12
6.
If the pH of a 0.10 M weak acid H2X is 3.683, calculate the Ka for the acid and identify the acid using your
acid chart. (the first H+ donation is the only significant one)
pH = 3.683 some [H3O+] = 2.07 X 10-4 Ka = X2/ (0.10-2.07 X 10-4 ) which is X
Ka = (2.07 X 10-4 )2 / (0.10- 2.07 X 10-4 )
Ka = 4.31 X 10-7 probably Carbonic acid as it’s Ka is = 4.4 X 10-7
7.
The pH of 0.65 M NaX is 12.46. Calculate the Kb for NaX.
pH = 12.46, so pOH = 1.54 some [OH-] = 0.0288 which is X
Kb = X2/ (0.65 -2.07 X 10-4 ) or
Kb = (0.0288 )2 / (0.65 - 0.0288 )
Kb = 1.34 X 10-3 probably Carbonic acid as it’s Ka is = 4.4 X 10-7
Ka = (1.00 x 10-14/ Kb X- (aq) 1.34 X 10-3 )
Ka = 7.5x10-12
8.
Consider the following reaction: 2HCl + Ba(OH)2 → BaCl2 + 2H2O
When 3.16g samples of Ba(OH)2 were titrated to the equivalence point with an HCl solution, the following
data was recorded. Calculate the original [HCl]
Trial
Volume of HCl added
#1
37.80 mL
#2
35.49 mL
#3
35.51 mL
Answer to #8: Find moles of 3.16g of Ba(OH)2 / 171.35 = 0.0184 moles and X by 2/1 which is moles of
acid = 0.0369 moles and C=n/v (where volume is average of last 2 trials of HCL V= 35.50 ml
C= 0.0369 mol/ 0.03550L = [1.04] is Conc of HCl
9.
Calculate the volume of 0.200M H2SO4 required to neutralize 25.0 ml of 0.100M NaOH.
2NaOH + H2SO4 → Na2SO4 + 2H2O
nNaOH= CV (0.100mol/l)(0.02500 L) = 0.00250 mole of base x 1/2 = 0.00125 moles acid reacted
V=n/C = 0.00125 mol / 0.200mol/l = 0.00625 L or 6.25 ml
10.
25.0 mL of 0.200 M HCl is mixed with 50.0 mL 0.100 M NaOH, calculate the pH of he resulting solution.
1:1 mole ratio for reaction: HCl + NaOH → NaCl + H2O
or (HOH)
nHCl= CV (0.200mol/l)(0.0250 L) = 0.00500mole of acid
nNaOH= CV (0.100mol/l)(0.0500 L) = 0.00500mole of base, so moles are the SAME and will cancel.
Thus we are left with only the products, no excess reactants. The products are water and salt, so pH =7.00
11.
125.0 mL of 0.200 M HCl is mixed with 350.0 mL 0.100 M NaOH, calculate the final pH.
HCl + NaOH → NaCl + H2O
or (HOH)
nHCl= CV (0.200mol/l)(0.1250 L) = 0.02500mole of acid
nNaOH= CV (0.100mol/l)(0.3500 L)= 0.03500mole base, we have EXCESS base (0.03500mole-0.02500mole)
C = nVtotal 0.0100 mol/ 0.475L = 0.0210 M final excess {OH-] so pOH = 1.68 and pH = 12.32
12.
What is the [H3O+] in a solution formed by adding 60.0 mL of water to 40.0 mL of 0.040 M KOH solution?
This is weird as we have base being diluted NOT reacted so new conc
Ci
Vi
=
Cf
Vf
0.040Mx0.040 L
= Cf x 0.0600L
0.02666 M
= Cf of BASE
pOH = 1.57 and pH =12.43
13.
Out of 0.2 M HCl and 1.0 M HF, which is the most concentrated? Which is the strongest acid, and why?
HCl reacts 100% WHEREAS THE WEAKER HF REACTS 8%, BUT HCL IS LESSS CONCENTRATED
14.
What volume of 0.200 M NaOH is required to neutralize 25.0 mL of 0.150 M H2SO4?
2NaOH + H2SO4 → Na2SO4 + 2H2O
n H2SO4 = CV (0.250mol/l)(0.1500 L) = 0.0375 mole of ACID x 2/1 = 0.0750 moles base
V=n/C = 0.0750mol / 0.200mol/l = 0.0375 L or 37.5 ml
15.
In a titration 25.0 mL of 0.200M H2SO4 is required to neutralize 10.0 mL NaOH. Calculate the
concentration of the base. 2NaOH + H2SO4 → Na2SO4 + 2H2O
n H2SO4 = CV (0.200mol/l)(0.2500 L) = 0.0500 mole of ACID x 2/1 = 0.100 moles base
C=n/v = 0.10mol / 0.100 L = 1.00 M or 1.00 mol/L Base NaOH