Question

9. 15.00 mL of 0.100 M NaOH are added to 50.00 mL of a buffer solution containing 1.0 mmol CH3COOH and 1.0 mmol CH3C00-. What is the new solution pH? (a) 11.89 (b) 2.11 (c) 7.87 (d) Can't tell / need more info. 10. The equilibrium constant for the below reaction is 4.2 x 10 Which of the following statements is FALSE? CuBr (s) = Cu+ (aq) + Br- (aq) (a) CuBr has poor solubility in water. (b) If 15.0 g CuBr (s) are added to a solution saturated in Cut and Br", more CuBr will dissolve. (c) At equilibrium, the concentrations of Cut and Br" will be equal. (d) AG xn is positive. 11. Buffer X contains 8.0 mmol of NH, and 8.0 mmol of NH. Buffer Y contains 20.0 mmol of NH, and 20.0 mmol of NH. Which of the following statements is TRUE? (a) Buffer X will have a pH five times more than Buffer Y. (b) Buffer X will have a pH five times less than Buffer Y. (c) Buffer X and Buffer Y will have the same pH. (d) Buffer X and Buffer Y cannot be compared unless we have the pka of NHA.

Answer 1

Hello, i am sorry to say that as per HOMEWORKLIB RULES experts can answer only one full question that too first question only. I request you to please post remaining question as separate one since my allotted time could not permit me to answer more than one question also. Thank you in advance for your continued support to us.

1)

moles OH^- in NaOH = 15.00 mL x 10^-3 L x 0.100 M = 0.0015 mol

1 mmol = 0.001 mol

moles CH₃COOH = 50 mL x 1.0 mmol x 0.001 = 0.05 mol

moles CH₃COO^- = 50 mL x 1.0 mmol x 0.001 = 0.05 mol

Total volume = 15 mL + 50 mL = 65 mL = 0.065 L

[OH-] = 0.0015 mol / 0.065 L = 0.0231 M

[CH₃COOH] = 0.05 mol / 0.065 L = 0.769 M

[CH₃COO^-] = 0.05 mol / 0.065 L = 0.769 M

R....................CH₃COOH..........+..........OH^-......<==>........CH₃COO^-

I........................0.769..........................0.0231.......................0.769

C.....................-0.0231......................-0.0231.......................0.0231

E.....................0.7459...........................0............................0.7921

CH₃COOH acid dissociation constant Ka = 1.8 x 10^-5

pKa = -log(Ka) = -log(1.8 x 10^-5) = 4.74

From Henderson-Hasselbalch equation for buffers

pH = pKa + log(conjugate base / acid)

pH = 4.74 + log[0.7921 / 0.7459]

pH = 4.77

New solution pH = 4.77

This option was not given in the question, so correct option is (d) can't tell/need more info

Hope this helped you!

Thank You So Much! Please Rate this answer as you wish.("Thumbs Up")