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.

Answer 1

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