1. How many grams of solid potassium cyanide should be added to 0.500 L of a 0.1

Question

1. How many grams of solid potassium cyanide should be added to 0.500 L of a0.134 M hydrocyanic acid solution to prepare a buffer with a pH of 8.530 ?

grams potassium cyanide = _____ g

2. A buffer solution contains 0.483 M KHSO3 and 0.304 M K2SO3.

If 0.0380 moles of potassium hydroxide are added to 225 mL of this buffer, what is the pH of the resulting solution?
(Assume that the volume does not change upon adding potassium hydroxide)

pH = ________

3. A buffer solution contains 0.480 M hydrocyanic acid and 0.391 M sodiumcyanide.
If 0.0330 moles of sodium hydroxide are added to 150 mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume does not change upon adding sodium hydroxide)

pH = ________

4. Determine the pH change when 0.113 mol KOH is added to 1.00 L of a buffer solution that is 0.448 M in HClO and 0.383 M in ClO-.
pH after addition pH before addition = pH change = _______

5. A buffer solution contains 0.380 M NH4Cl and 0.491 M NH3 (ammonia). Determine the pH change when 0.136 mol HBr is added to 1.00 L of the buffer.
pH after addition pH before addition = pH change = _____

6. Consider the insoluble compound iron(II) hydroxide , Fe(OH)2. The iron(II)ion also forms a complex with cyanide ions . Write a balanced net ionic equation to show why the solubility of Fe(OH)2(s) increases in the presence of cyanideions and calculate the equilibrium constant for this reaction.

For Fe(CN)64-, Kf = 7.7×1036. Use the pull-down boxes to specify states such as (aq) or (s).

K =

Explanation / Answer

1)

Ka of HCN = 6.2*10^-10

pKa = - log Ka

= -log (6.2*10^-10)

= 9.208

use:

pH = pKa + log {[KCN]/[HCN]}

8.530 = 9.208 + log {[KCN]/0.134}

log {[KCN]/0.134} = -0.678

[KCN]/0.134 = 0.2099

[KCN] = 0.0281 M

volume = 0.500 L

mol of KCN = [KCN]*volume

= 0.0281M * 0.500 L

= 0.01406 mol

molar mass of KCN = 65.12 g/mol

mass of KCN = number of mol * molar mass

= 0.01406 mol * 65.12 g/mol

= 0.916 g

Answer: 0.916 g

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