Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

9. Which of the following solutions has the greatest buffering capacity? [A] 0.2

ID: 966956 • Letter: 9

Question

9. Which of the following solutions has the greatest buffering capacity?

   [A] 0.243 M HF and 0.192 M NaF

   [B] 0.485 M HF and 0.383 M NaF

   [C] 0.970 M HF and 0.766 M NaF

   [D] 0.160 M HF and 0.126 M NaF  

   [E] They are all buffer solutions and would all have the same capacity.

   10. The Henderson-Hasselbalch equation is ________

   [A] pH = pKa + log [acid]/[base]

   [B] [H+] = Ka + [base]/[acid]

   [C] pH = log [acid]/[base]

   [D] pH = pKa - log [base]/[acid]

   [E] pH = pKa + log [base]/[acid]

7. Calculate the pH of a solution prepared by dissolving 0.150 mol of benzoic acid and 0.300

   mol of sodium benzoate in water sufficient to yield 1.00 L of solution. The Ka of benzoic

   acid is 6.30 x 10-5.

[A] 2.516 [B] 4.502 [C] 3.892 [D] 4.195 [E] 10.158

8. What change will be caused by addition of a small amount of HCl to a solution containing

fluoride ions and hydrogen fluoride?

[A] The fluoride ions will precipitate out of solution as its acid salt.

[B] The concentration of fluoride ion will decrease and the concentration of hydrogen

fluoride will increase.

[C] The concentration of hydronium ions will increase significantly

[D] The concentration of hydrogen fluoride will decrease and the concentration of

   fluoride ions will increase.

[E] The concentration of fluoride ions will increase as will the concentration of hydronium

   ions.

11. The addition of hydrofluoric acid and ________ to water produces a buffer solution

[A] HF [B] KI [C] NaF [D] NaBr [E] NaNO3

Explanation / Answer

9) when ratio of [NaF]/[HF] is near 1 , its buffer capacity is more

hence option A   since its ratio is nearer to 1 compared to all

10) pH = pka + log [base] /[acid]              ( base means conjugate base of acid)

7) pka = -log Ka = -log ( 6.3x10^-5) = 4.2

pH = pka + log [benzoate ]/[benzoic acid] = 4.2 + log ( 0.3/0.15)

= 4.5

8) H+ (aq) + F-(aq) ---> HF (aq)

thus concentration of Flouride ion decreases and concentration of hydrogen flouride increases

11) HF and NaF   makes buffer ( since we had acid HF and its conjugate base NaF)