PLEASEEEEE Determine the molarity of potassium hydroxide, KOH, given that 320.5

Question

PLEASEEEEE

Determine the molarity of potassium hydroxide, KOH, given that 320.5 mL of potassium hydroxide, KOH, needs 237.5 mL of 0.198 M sulfuric acid, H_2SO_4, to reach the equivalence point? What kind of problem do you think this is? 13) Calculate the pH of a solution prepared by mixing 100.0 mL of 0.178 M hydrochloric acid (HCl) with 75.0 mL of 0.185 M sodium hydroxide, NaOH. What kind of problem do you think this is? 14) Find the pH of a solution prepared by missing 120.0 mL of 0.254 M HCl with 240.00 mL of 0.127 M NaOH. What kind of problem do you think this is? 15) Calculate the pH of a solution prepared by mixing 100.0 mL of 0.178 M hydrochloric acid (HCl) with 100.0 mL of 0.185 M sodium hydroxide, NaOH. What kind of problem do you think this is? 16) Calculate the pH of a solution prepared by mixing 125.0 mL of 0.178 M acetic acid (HC_2H_3O_2) with 150.0 mL of 0.100 M sodium hydroxide, NaOH. What kind of problem do you think this is? 17) Calculate the pH of a solution prepared by mixing 125.0 mL of 0.178 M acetic acid (HC_2H_3O_2) with 222.5 mL of 0.100 M sodium hydroxide, NaOH. What kind of problem do you think this is? 18) Calculate the pH of a solution prepared by mixing 125.0 mL of 0.178 M acetic acid (HC_2H_3O_2) with 250.0 mL of 0.100 M sodium hydroxide, NaOH. What kind of problem do you think this is? 19) Consider the following equation 4(NH_4)_3PO_4 + 3 Pb(NO_3)_4 rightarrow Pb_3(PO_4)_4 + 12 NH_4NO_3 If (NH_4)3PO_4 disappears at a rate of 0.0012 Ms-1. During the same time period, determine the rate at which a) Pb(NO_3)_4 disappears b) Pb_3(PO_4)_4 appears c) NH_4NO_3 appears 20) Consider the following equation to complete the sentences below 4(NH_4)_3PO_4 + 3 Pb(NO_3)_4 rightarrow Pb_3(PO_4)_4 + 12 NH_4NO_3 a) Pb(NO_3)_4 disappears times as fast as NH_4NO_3 appears b) Pb(NO_3)_4 disappears times as fast as Pb_3(PO_4)_4 appears c) (NH_4)_3PO_4 disappears times as fast as Pb(NO_3)_4 disappears 21) By what factor will the overall rate increase when the concentration of A is increased five times, concentration of B is quadrupled and the concentration of C is cut in half? The rate law of the reaction is Rate = k [A]^3[B]^1/2[C]^4

Explanation / Answer

12. moles of KOH = 1/2 moles of H2SO4

                             = 1/2(0.198 M x 237.5 ml)

                             = 47.025 mmol

concentration of KOH = 47.025 mmol/320.5 ml = 0.147 M

This is neutralization reaction

13. moles of HCl = 0.178 M x 100 ml = 17.8 mmol

moles of NaOH = 0.185 M x 75 ml = 13.875 mmol

excess [H+] = 3.925 mmol/175 ml = 0.022 M

pH = -log[H+] = 1.65

This is neutralization reaction. Strong acid-strong base titration

14. moles of HCl = 0.254 M x 120 ml = 30.48 mmol

moles of NaOH = 0.127 M x 240 ml = 30.48 mmol

pH = 7

This is neutralization reaction. Strong acid-strong base titration

15. moles of HCl = 0.178 M x 100 ml = 17.8 mmol

moles of NaOH = 0.185 M x 100 ml = 18.5 mmol

excess [OH-] = 0.7 mmol/200 ml = 0.0035 M

pOH = -log[OH-] = 2.456

pH = 11.544

This is neutralization reaction. Strong acid-strong base titration

16. moles of CH3COOH = 0.178 M x 125 ml = 22.25 mmol

moles of NaOH = 0.1 M x 150 ml = 15 mmol

[CH3COOH] remined = 7.25 mmol/275 ml = 0.026 M

[CH3COO-] formed = 15 mmol/275 ml = 0.054 M

pH = pKa + log(base/acid)

     = 4.74 + log(0.054/0.026)

     = 5.06

This is preparation of buffer solution

17. moles of CH3COOH = 0.178 M x 125 ml = 22.25 mmol

moles of NaOH = 0.1 M x 222.5 ml = 22.25 mmol

[CH3COO-] formed = 22.25 mmol/347.5 ml = 0.064 M

CH3COO- + H2O <==> CH3COOH + OH-

Kb = 1 x 10^-14/1.8 x 10^-5 = x^2/0.064

x = [OH-] = 5.96 x 10^-6 M

pOH = -log[OH-] = 5.22

pH = 14 - pOH = 8.78

This is weak acid-strong base titration

18. moles of CH3COOH = 0.178 M x 125 ml = 22.25 mmol

moles of NaOH = 0.1 M x 250 ml = 25 mmol

excess [OH-] = 2.5 mmol/375 ml = 0.0067 M

pOH = -log[OH-] = 2.17

pH = 14 - pOH = 11.83

This is titration od weak acid-strong base.

19. For the given reaction,

rate of dissappearence of (NH4)3PO4 = 0.0012 M/s

so rate of,

a) dissappearence of Pb(NO3)4 = 0.0012 x 3/4 = 0.0009 M/s

b) appearence of Pb3(PO4)4 = 0.0012/4 = 0.0003 M/s

c) appearence of NH4NO3 = 0.0012 x 12/4 = 0.0036 M/s

20. For the reaction,

a) Pb(NO3)4 disappears 0.25 times as fast as NH4NO3 appears

b) Pb(NO3)4 disappears 3 times as fast as Pb3(PO4)4 appears

c) (NH4)3PO4 disappears 1.33 times as fast as Pb(NO3)4 disappears

22. By increasing A by fives time, B by 4 times and reducing C by half, the rate would increase by a factor of 15.625 times

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