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I would really appreciate it if someone showed me how to solve each of these pro

ID: 869982 • Letter: I

Question

I would really appreciate it if someone showed me how to solve each of these problems instead of just giving the answers. I bolded the correct answer. Thank you

1. How many milliliters of a 5.0 x 10^-2 stock solution would you need to add to a final volume of 10 liters to obtain a solution with a pH of 3.60?

a. 50 mL

b. 34 mL

c. 13 mL

2. A 0.25 M solution of a hypothetical weak acid, HX has a pH of 3.42. What is the Ka value for this acid?

a. 8.8 x 10^ -6

b. 1.6 x 10^ -7

c. 5.8 x 10^ -7

3. What is the pH of a 0.30 M solution of magnesium hypochlorite(Mg(ClO)2)? For HClO Ka= 3.0 x10^ -8

a. 9.06

b. 3.06

c. 10.65

4. The pH= 2.84 for a 0.35 M pyridine hydrochoride, C5H5NHCl, solution. Based on this information, determine the Kb value of pyridine (C5H5N)

a. 9.4 x10^-8

b. 1.7x 10^ -9

c. 5.3 x 10^ -7

5. If you wanted to prepare a buffer composed of hypochlorous acid (HClO0 and potassium hypochlorite (KClO), how may moles of KClO would be added to 2 liters of a 1.0 M HClO solution to get a buffer with a final pH of 7.70? For HClO, Ka= 3.0 x 10^-8

a. 3.0 mol KClO

b. 1.2 mol KClO

c. 2.2 mol KClO

6. The pH of a solution by mixing 300 mL of 0.050 M HClO4 and 40.0 mL of 0.10 M Ba(OH)2 is?

a. 1.69

b. 1.77

c. 1.85

7. A 150 mL sample of 0.20 M HF solution is titrated with a strong base. What is the pH after adding 40 mL of 0.80 M NaOH? For HF, Ka= 6.8 x 10^-4

a. 13.11

b. 12.02

c. 10.37

8. A solution of Pb(NO3)2 is mixed with a solution of NaCl. A precipitate forms and analysis of the supernatant reveals the following concentrations of ions.

[Pb^2+] = 0.0062 M [NO3-]= 0.25 M [Na+]= 0.40 M [Cl-]= 0.054 M

Based on these data, what is the Ksp value for PbCl2(s)?

a. 7.2 x 10^-6

b. 5.6 x 10^-8

c. 1.8 x10^ -5

9. At an elevated temperature, Ksp= 7.5x 10^ -5 for Ca(OH)2. What is the pH of a saturated solution of a calcium hydroxide?

a. 13.06

b. 9.51

c. 12.73

Explanation / Answer

1)

V1 =

M1 = 5.0 x 10^-2

V2 = 3.60 L

M2 = ?

get M2 with pH = 3.6

[H+] = 10^-ph = 10^-3.6 = 0.0002512

M2 = 0.0002512

M1*V1 = M2*V2

5*10^-2 * V1 = 0.0002512*10

V1 = 0.05 liters or 50 ml

2)A 0.25 M solution of a hypothetical weak acid, HX has a pH of 3.42. What is the Ka value for this acid?

M1 = 0.25 M HX

ph = 3.42

[H+] = 10^-3.42 = 0.00038

Ka = [H+][X-] / [HX]

[H+] = [X-] = 0.00038

[HX] = 0.25 -0.00038 = 0.2496

Ka = [H+][X-] / [HX] = (0.00038*0.00038) / (0.2496) = 5.78*10^-7

3) What is the pH of a 0.30 M solution of magnesium hypochlorite(Mg(ClO)2)? For HClO Ka= 3.0 x10^ -8

Kb = Kw/Ka = (10^-14)/(3*10^-8) = 3.33*10^-7

Assume all MgClO2 are in solution, then 0.30 mol will give 2*0.3 mol of ClO-

0.6 M of ClO-

ClO- + H2O <--> HClO + OH-

Kb = [OH-][HClO] / [ClO-]

[OH-] = [HClO] = x

[ClO-] = 0.6

Solve for x

3.33*10^-7 = x*x / (0.6)

x^2 = 3.33*10^-7 * 0.6 = 1.99*10^-7

x = sqrt ( 1.99*10^-7 ) = 0.000446 OH-

pOH = -log(Oh-) = -log(0.000446) = 3.35

pH = 14- pOH = 14-3.35 = 10.65

4) The pH= 2.84 for a 0.35 M pyridine hydrochoride, C5H5NHCl, solution. Based on this information, determine the Kb value of pyridine (C5H5N)

Pyridine Kb value is actually a reported value in books, this is a constant and will not change (if temperature is not changed)

PLEASE Consider posting all other questions in another set of Q&A