****calculate the pH of a 0.5M solution of KCN. ****how do contentraion/volumes

Question

****calculate the pH of a 0.5M solution of KCN.

****how do contentraion/volumes of the buffer affect the buffer capcaity

Chemistry 106. 106 Name: Post-Lab Assessment Questions 020 points): tration and p of 0-8? monia 3. (4 pts) Calculate the pH of a 0.500 M solution of KCN. Ka for HCN is 5.8 x 10-10. 4. (8 pts) How do the concentration/volumes of the buffer affect the buffer capacity? E,g, 50.0 mL of 0.10 M acetic acid solution with 50.0 mL of 0.10 M sodium acetate solution vs. the buffer you made in the lab (25.0 mL of 0.10 M acetic acid solution with 25.0 mL of 0.10 M sodium acetate solution)

Explanation / Answer

3)

CN- is the conjugate base of the weak acid HCN (Ka 5.8 X 10^-10) therefore CN-1 is a strong base

CN is strong base, so find Kb for HCN

Kb = Kw/Ka

Kb = 1 x 10^-14/5.8 x 10^-10

Kb = 1.724 x 10^-5

KCN will hydrolyze in the following way

CN-1 + H2O ----> HCN + OH-

I 0.500                 0          0

C –x                     x           x

E 0.500-x            x            x

Kb = [HCN][OH]/[CN]

1.724 x 10^-5 =[ x][x]/0.500-x

x will be very small compared to 0.500 so neglect it in the denominator

x^2 = 0.500 x 1.724 X 10^-5

x^2= 8.62 X 10^-6

x = 2.94 X 10^-3 = [OH]

pOH = -log (2.94 X 10^-3 ) = 2.53

PH = 14 – POH

PH = 14 – 2.53 = 11.47

pH = 11.47

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