Question

Buffers
Q1:
a. Write the balanced chemical equation for the reaction of HCN with water.

answer: HCN + H2 O----> CN- + H3O
b. What is the pH and concentration of CN– in a 0.100M solution of HCN?

answer: [CN-]=7.87 x 10-6     pH: 5.1

c. What would happen if the equilibrium [CN–] that you calculated in (b) was changed by adding 0.00500 moles of solid NaCN to 100.0 mL of the solution? (Which direction is the reaction going to “go” to minimize the disturbance?)

answer: Backward direction

d. Determine the pH of the solution in (c) . Assume that the addition of solid NaCN does not change the volume of solution.

answer :pH:8.91

e. What is the name of this type of solution?

answer: Buffers

f. Determine the pH of the solution from (c) AFTER 10.0 mL of 0.100M HNO3 is added.
g. Determine the pH of the solution from (c) AFTER 10.0 mL of 0.100M KOH is added.
h. How many moles of strong acid could you reasonably expect to add to the solution in (c) before the pH would change drastically?
i. How many moles of strong base could you reasonably expect to add to the solution in (c) before the pH would change drastically?

Answer 1

Q1:

(a).

The balanced chemical equation for the reaction of HCN with water is

HCN + H₂O CN^- + H₃O⁺

(b)

HCN + H₂O    CN^- + H₃O⁺

[HCN] = 0.1 M

Ka of HCN = 6.2 x 10^-10

Now, Ka = [CN^-] [H₃O⁺] / [HCN]

6.2 x 10^-10 = x² / 0.1

x² = 6.2 x 10^-11

x = 7.87 x 10^-6

So, = [H₃O⁺] = x = 7.87 x 10^-6

pH = -log [H₃O⁺] = - log (7.87 x 10^-6) = 5.10

(c)

HCN + H₂O    CN^- + H₃O⁺

When 0.005 moles of solid NaCN added;

NaCN is an salt and dissociate completely. So, moles of NaCN is equal to moles of CN^-. Now, when 0.005 moles of CN^- (product added) added to the equilibirium, the equilibirium will shift to the reactant side or left direction (or backward direction).

(e)

HCN + H₂O    CN^- + H₃O⁺

HCN is a weak acid and CN^- is the corresponding conjugated base.

So its a Buffer solution.