Question

Which solutions below would be described as a buffer solution?

A) 500 mL of 0.1 M HCl added to 500 mL of 0.1 M NaOH

B) 500 mL of 0.1 M HCl added to 500 mL of 0.1 M NaCl

C) 500 mL of 0.1 M HCNO added to 500 mL of 0.1 M KCNO

D) 500 mL of 0.1 M HCNO added to 250 mL of 0.1 M KCNO

E) 500 mL of 0.1 M HCNO added to 250 mL of 0.1 M NaOH

F) 500 mL of 0.1 M HCNO added to 500 mL of 0.1 M NaOH

2. Determine the pH of the solution in question 1E given the K_a(HCNO) = 2 ´ 10^‒4.

Answer 1

1) answers are

C) 500 mL of 0.1 M HCNO added to 500 mL of 0.1 M KCNO

D) 500 mL of 0.1 M HCNO added to 250 mL of 0.1 M KCNO

E) 500 mL of 0.1 M HCNO added to 250 mL of 0.1 M NaOH

2)

Ka(HCNO) = 2 ´ 10‒4.

pKa = - log Ka = - log (2 ´* 10^‒4.) = 3.70

500 mL of 0.1 M HCNO = 0.500 L * 0.1 mole / L = 0.050 mole.

and

mole of NaOH = 0.250 L * 0.1 mole / L = 0.025 mole.

HCNO + NaOH .............> NaCNO + H2O

mole of NaCNO = 0.025 mole.

and

mole of HCNO = (0.050 - 0.025) = 0.025 mole.

Using Henderson equation,

pH = pKa + log [salt] / [acid]

or

pH = 3.70 + log (0.025 / 0.025)

or

pH = 3.70 (answer)