1) You are asked to prepare 2.8 L of a HCN/NaCNbuffer that has a pH of 9.85 and

ID: 529101 • Letter: 1

Question

1) You are asked to prepare 2.8 L of a HCN/NaCNbuffer that has a pH of 9.85 and an osmotic pressure of 1.73 atm at 298 K. What masses of HCN and NaCN should you use to prepare the buffer? (Assume complete dissociation of NaCN.)

2) Find the mass of sodium formate that must be dissolved in 290.0 cm3 of a 0.50 M solution of formic acid to prepare a buffer solution with pH = 3.60.

3) Two 18.0 mL samples, one 0.200 MKOH and the other 0.200 M CH3NH2, were titrated with 0.100 MHI. Answer each of the following questions regarding these two titrations.

3.a) What is the volume of added acid at the equivalence point for KOH?

3.b) What is the volume of added acid at the equivalence point for CH3NH2?

3.c)Predict whether the pH at the equivalence point for each titration will be acidic, basic, or neutral.

Predict whether the at the equivalence point for each titration will be acidic, basic, or neutral.

Predict which titration curve will have the lowest initial pH.

Predict which titration curve will have the lowest initial .

basic for KOH and neutral for CH3NH2 acidic for KOH and neutral for CH3NH2 neutral for KOH and basic for CH3NH2 neutral for KOH and acidic for CH3NH2 neutral for both

Explanation / Answer

Answer to Q1)

The osmotic pressure for a solution is given by the equation:

= M R T

where, M = molar concentration of dissolved species

R = ideal gas constant = 0.08206 L atm mol-1 K-1

T = temperature in Kelvin

M = / R T

= 1.73 atm / 0.08206 L atm mol-1 K-1 • 298 K

= 0.0707 mol L-1 = 0.0707 M

pH = -log [H+]

[H+] = 10-pH

= 10-9.85

= 1.412 X 10-10 M

HCN is a weak acid. It dissociates as:

HCN <-------> H+ + CN-

Initially CN- and HCN are present which can be calculated using ICE table

[H+]

[CN-]

[HCN]

Initial

x M

y M

Change

+1.412 X 10-10 M

+ 1.412 X 10-10 M

-1.412 X 10-10 M

Equilibrium

1.412 X 10-10 M

0.0707 M

Since the concentration of H+ ion is very small compared to the concentration of HCN and CN- ions in the solution, we can assume

[CN-] = x = 0.0707 M - 1.412 X 10-10 M = 0.0707 M

[HCN] = y = 0.0707 M + 1.412 X 10-10 M = 0.0707 M

Number of moles of NaCN = [CN-] X volume of the solution

= 0.0707 M X 2.8 L

= 0.19796 mol

Mass of NaCN = Number of moles of NaCN X molar mass of NaCN

= 0.19796 mol X 49.0072 g mol-1

= 9.7015 g

Number of moles of HCN = [HCN] X volume of the solution

= 0.0707 M X 2.8 L

= 0.19796 mol

Mass of HCN = Number of moles of HCN X molar mass of HCN

= 0.19796 mol X 27.0253 g mol-1

= 5.3499 g