Calculate the volume of 0.500 M C2H3O2H and 0.500 M C2H3O2Na required to prepare 0.100 L of pH 5.00 buffer with a buffer strength of 0.100 M. The pKa of C2H3O2H is 4.75.

You require 12.8mL of the 0.500M C₂H₃O₂Na and 7.2mL of the 0.500M C₂H₃O₂H

Explanation:

It is possible to obtain pH of a weak acid using H-H equation:

pH = pKa + log₁₀ [A⁻] / [HA]

For the buffer of acetic acid/acetate, the equation is:

pH = pKa + log₁₀ [C₂H₃O₂Na] / [C₂H₃O₂H]

Replacing:

5.00 = 4.75 + log₁₀ [C₂H₃O₂Na] / [C₂H₃O₂H]

1.7783 =  [C₂H₃O₂Na] / [C₂H₃O₂H] (1)

Buffer strength is the concentration of the buffer, that means:

0.1M = [C₂H₃O₂Na] + [C₂H₃O₂H] (2)

Replacing (2) in (1):

1.7783 =  0.1M - [C₂H₃O₂H] / [C₂H₃O₂H]

1.7783 [C₂H₃O₂H] = 0.1M -  [C₂H₃O₂H]

2.7783 [C₂H₃O₂H] = 0.1M

Also:

[C₂H₃O₂Na] = 0.1M - 0.036M

The moles of both compounds you require is:

[C₂H₃O₂Na] = 0.1L × (0.064mol / L) = 0.0064moles

[C₂H₃O₂H] = 0.1L × (0.036mol / L) = 0.0036moles

Your stock solutions are 0.500M, thus, volume of both solutions you require is:

[C₂H₃O₂Na] = 0.0064moles × (1L / 0.500M) = 0.0128L = 12.8mL

[C₂H₃O₂H] = 0.0036moles × (1L / 0.500M) = 0.0072mL = 7.2mL

the maximum mass of nh3 is nh3 in grams

yes the answer is total mass in unchanged   : )