1) Calculate of pH of each of the following strong acid solutions: (a) 8.5 x 10^
ID: 949737 • Letter: 1
Question
1) Calculate of pH of each of the following strong acid solutions: (a) 8.5 x 10^-3 M HBr, (b) 1.52 g of HNO3 in 575 mL of solution, (c) 5.00 mL of 0.250 M HClO4 diluted to 50.0 mL, (d) a solution formed by mixing 10.0 mL of 0.100 M HBr with 20.0 mL of 0.200 M HCl.
2) Calculate [OH-] and pH for (a) 1.5 x 10^-3 M Sr(OH)2, (b) 2.250 g of LiOH in 250.0 mL of solution, (c) 1.00 mL of 0.175 M NaOH diluted to 2.00 L, (d) a solution formed by adding 5.00 mL of 0.105 M KOH to 15.0 mL of 9.5 x 10^-2 M Ca(OH)2
3) Latic acid (CH3CH(OH)COOH) has one acidic hydrogen. A 0.10 M solution of lactic acid has a pH of 2.44. Calculate Ka.
4) A 0.100 M solution of chloroacetic acid (ClCH2COOH) is 11.0% ionized. Using this information, calculate [ClCH2COO-], [H+], [ClCH2COOH], and Ka for chloroancetic acid.
Explanation / Answer
Answer for question (1a):
pH of the following strong acid solution:
pH = -log [H+] = - log (8.5 x 10-3) = 2.0705
Answer for question (1b):
Molarity of HNO3 solution = No. of moles/volume in L
No. of moles HNO3 = 1.52 g / 63.01 = 0.0241 Moles
volume in L = 575 ml / 1000 ml = 0.575 L
Molarity of HNO3 solution = 0.0241/ 0.575 = 0.0419 M
So [H+] = 0.0419 M
pH = - log (0.0419) = 1.377
Answer for question (1c):
HClO4 Molarity, M1 = 0.250 M
Volume of HClO4, V1 = 5.0 ml
Diluted to volume, V2 = 50.0 mL
M1V1 = M2V2
M2 = 0.25 x 5 / 50 = 0.025 M
[H+] = 0.025 M
pH = - log (0.025) = 1.602
Answer for question (1d):
Moles of HBr = 10 x 0.1 = 1.0 mMole = 0.001 Mole
Moles of HCl = 20 x 0.2 = 4.0 mMole = 0.004 Mole
Total No. of combined moles of H+ = 1 + 4 = 0.005 Moles
Total volume of solution = 30 ml = 0.03 L
Molarity of mixsd acid solution = 0.005/0.03 = 0.1666 M
pH = - log (0.1666) = 0.7781
Answer for question (2a):
Sr(OH)2 <=====> Sr2+ + 2 OH-
1 mole of Sr(OH)2 gives 2 moles of OH-
So 1.5 x 10-3 M Sr(OH)2 gives 3.0 x 10-3 Moles of OH-
[ OH-] = 3.0 x 10-3 Moles
pOH = -log (3.0 x 10-3) = 2.52
Sr(OH)2 pH = 14 - pOH = 14 - 2.52 = 11.48
Answer for question (2b):
2.250 g of LiOH in 250.0 mL of solution
No. of moles of LiOH = 2.250/23.95 = 0.0939 Moles
Volume in litres = 250/1000 = 0.25 L
Molarity of LiOH solution = 0.0939/0.25 = 0.3756
[OH-] = 0.3756 M
pOH = -log (0.3756) = 0.425
LiOH solution pH = 14- pOH = 14 - 0.425 = 13.575
Answer for question (2c):
1.00 mL of 0.175 M NaOH diluted to 2.00 L
Initial Molarity of NaOH, M1= 0.175M
Initial volume of NaOH solution = 1.0 ml
Final volume of NaOH solution = 2.0 ml
Final Molarity of NaOH, M2 = 1 x 0.175 / 2 = 0.0875 M
[OH-] = 0.0875 M
pOH = -log (0.0875) = 1.058
pH of NaOH solution = 14 - 1.058 = 12.942
Answer for question (2d):
5.00 mL of 0.105 M KOH to 15.0 mL of 9.5 x 10^-2 M Ca(OH)2
Moles of KOH = 5 x 0.105 = 0.525 mMoles = 0.000525 Moles
Moles of Ca(OH)2 = 2 x 9.5 x 10-2 = 15 x 0.19 mMoles = 2.85 mMoles = 0.00285 Moles
Mixed moles of [OH] in combined solution = 0.000525 Moles + 0.00285 Moles = 0.003375 Moles
Total volume = 15 + 5 = 20 ml = 0.02 L
Molarity of mixed solution = 0.003375/0.02 = 0.16875
pOH = -log 0.16875 = 0.7727
pH of mixed solution = 14 - 0.7727 = 13.2273
Answer for question (3):
Lactic acid (HA) dissociates into H+ and A- and their concentrations are as follows
[HA] = 0.1 M
pH = 2.44
Therefore concentration of H+ = 10-pH = 10-2.44 = 0.00363 M
Since HA dissociates into H+ and A- and their concentrations will be equal in strength
[H+] = [A-] = 0.00363M
Ka = [H+] [A-] / [HA] = 0.00363 x 0.00363 / 0.1 = 1.317 x 10-4
Lactic acid Ka = 1.317 x 10-4
Answer for question (4):
0.100 M solution of chloroacetic acid (ClCH2COOH) = 0.100 M
Dissociation of chloroacetic acid is 11.0%
Initial ClCH2COOH concentration = 0.1 M
% H+ = 11 Molar % = 0.011 M
ClCH2COO- concentration = 11 Molar% = 0.011 M
Unionized HA= 0.100 - 0.011 = 0.089 M
Ka = [H+] [ClCH2COO-]/[HA = 0.011 x 0.011 / 0.089 = 0.001359 = 1.359 x 10-3
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