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Worksheet: Acid base problems - AP level
Problems 1 - 10
Problem #1: Calculate the pH of the solution that results when 40.0 mL of 0.100 M NH 3 is:
a) Diluted with 20.0 mL of distilled water. b) Mixed with 20.0 mL of 0.200 M HCl solution. c) Mixed with 20.0 mL of 0.200 M NH 4 Cl
Solution to part (a):
- Use the dilution equation:
M 1 V 1 = M 2 V 2
(0.100 mol/L) (0.040 L) = (x) (0.060 L)
x = 0.0667 M
- Use the Kb for ammonia to determine pH:
Kb = ([NH 4 +] [OH¯]) / [NH 3 ]
1.77 x 10¯^5 = (x) (x) / (0.0667 - x)
Remember to ignore the x in 0.0667 - x.
x = 0.001086278 M (a few guard digits)
pOH = - log 0.001086278 = 2.
pH = 11.
Solution to part (b) and (c): Someday!
Problem #2: How many grams of ammonia are needed to make 1.25 L solution with a pH of 11.68?
Solution:
- Use the pH to get the hydroxide ion concentration:
pH = 11.
pOH = 14.00 - 11.68 = 2.
[OH¯] = 10¯pOH^ = 10¯2.
[OH¯] = 4.7863 x 10¯^3
- Use the Kb expression to get the [NH 3 ]:
Kb = ([NH 4 +] [OH¯]) / [NH 3 ]
1.77 x 10¯^5 = (4.7863 x 10¯^3 ) (4.7863 x 10¯^3 ) / (x)
x = 1.294275 M (guard digits!)
- Grams needed for 1.25 L
MV = g/molar mass
(1.294275 mol/L) (1.25 L) = x/17.031 g/mol
x = 27.6 g
Problem #3: Calculate the pH of a formic acid solution that contains 1.45% formic acid by mass. (Assume a density of 1.01 g/mL for the solution.)
Solution:
- Calculate volume of 100.0 g of solution:
100.0 g divided by 1.01 g/mL = 99.01 mL
- Calculate molarity of 1.45% solution of HCOOH:
1.45 g divided by 46.025 g/mol = 0.0315 mol
Problem #5: A buffer solution contains 0.348 M ammonium chloride and 0.339 M ammonia. If 0.0248 moles of hydrochloric acid are added to 125.0 mL of this buffer, what is the pH of the resulting solution? (Assume that the volume does not change upon adding hydrochloric acid.)
Solution:
A solution on video is provided for this problem.
Problem #6: How many mL of 4.50 M sodium hydroxide must be added to 250.0 mL of a 0.200 M acetic acid solution to make a buffer with pH = 5.000?
Solution:
- Use H-H Equation to determine required ratio of acetate to acid in solution:
5.000 = 4.752 + log [base] /[acid]
log [base] /[acid] = 0.
[base] /[acid] = 1.
- Determine molar amount of base required to get pH = 5.000 (for convenience, I'm going to use 1.00 L. I'll go to 250 mL at the end of this step):
1.77x + x = 0.
x = 0.0722 mol (this is acetic acid needed in the solution)
0.200 - 0.0722 = 0.1278 mol of base required
0.1278 /4 = 0.03195 mol of acetate required in 250 mL
- Determine volume of NaOH solution required:
4.50 mol/L = 0.03195 mol / x
x = 0.0071 L = 7.1 mL
- Check everything:
[acetic acid] = 0.01805 mol / 0.2571 L = 0.070206 M
[acetate] = 0.03195 mol / 0.2571 L = 0.12427 M
pH = 4.752 + log (0.12427 / 0.070206)
pH = 4.752 + log 1.
pH = 4.752 + 0.248 = 5.
Problem #7: If in a solution 10[H 3 O+] = [OH¯] then the pH of the solution is:
Solution:
- Let:
[H 3 O+] = x [OH¯] = y
- Therefore, two equations in two unknowns:
10x = y
and
xy = 1.00 x 10¯^14
- Rearrange the second equation:
y = (1.00 x 10¯^14 ) / x
- Substitute into the first equation:
10x = (1.00 x 10¯^14 ) / x
- Rearrange and solve:
10x^2 = 1.00 x 10¯^14
x^2 = 1.00 x 10¯^15
- Determine moles of oxalic acid that reacted with excess NaOH:
(0.0525 mol/L) (0.00995 L) = 5.22375 x 10¯^4 mol
- Determine moles of unreacted NaOH:
This reaction: H 2 C 2 O 4 (aq) + 2NaOH(aq) ---> Na 2 C 2 O 4 + 2H 2 O(l)
demonstrates a 1:2 molar ratio between oxalic acid and NaOH.
Therefore, we multiply the oxalic acid amount by two:
5.22375 x 10¯^4 mol x 2 = 1.04475 x 10¯^3 mol
to determine the amount of NaOH that did not react with the ammonium bromide.
- Determine moles of NaOH that did react with NH 4 Br:
5.125 x 10¯^3 mol minus 1.04475 x 10¯^3 mol = 4.08025 x 10¯^3 mol
- Determine grams of NH 4 Br:
4.08025 x 10¯^3 mol times 97.943 g/mol = 0.399632 g
To three sig figs, this is 0.400 g
Example #10: If 0.50 moles Ca(OH) 2 is slurried in 0.50 L deionized water and treated with 0.50 moles of CO 2 gas in a closed system, the liquid phase of this system will have a pH closest to what value?
Solution:
After the Ca(OH) 2 and the CO 2 react, we are left with some calcium carbonate, an insoluble substance. However, from the Ksp of CaCO 3 , we can calculate the approximate molarity of carbonate in the aqueous phase. Please see here for a discussion:
I will use the 5.5 x 10-5^ M from the above link.
Carbonate is the salt of a weak acid and so it hydrolyzes in solution:
CO 3 2-^ + H 2 O <==> HCO 3 -^ + OH-
To describe that system, we require the Kb1 of carbonate, which we get from the Ka2 of carbonic acid, which is 4.7 x 10-11, from here.
So the Kb1 of carbonate is 2.13 x 10-4^ (from KaKb = Kw)
We can now calculate the [OH-] in our calcium carbonate solution:
[OH-] = SQRT[(2.13 x 10-4) (5.5 x 10-5)] = 0.000108 M
The pOH is just under 4, which makes the pH be just over 10.
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