Question

5. How many grams of solid NaCNO must be added to 100 mL of 0.2 M HCNO at 25°C to obtain a buffer at pH = 4.00?

13. What is the pH of the buffer formed when 200 mL of 0.04 M propionic acid is mixed with 50 mL of 0.12 M sodium propionate (NaC₃H₅O₂)at 25 °C?

Answer 1

5.

volume of HCNO = 100.0 mL = 0.100 L
[HCNO] = 0.200 M

[CNO^-] = ? to have a buffer solution with pH = 4.00

We can use the Henderson-Hasselbalch equation to solve this problem.
pH = pKa + log{[CNO^-]/[ HCNO]}
pKa [HCNO] = 3.45

therefore;

pH = pKa + log{[CNO^-] /[ HCNO]}

4.00 = 3.45 + log[CNO^-] - log[0.2]

0.55= log[CNO^-] – (- 0.698)

log[CNO^-] = –0.148

[CNO^-] =10^ - 0.148

[CNO^-] =0.711

Check: 4.00 = 3.45 + log(0.711)/(0.200)

= 3.45 + 0.55

= 4.00

Hence; the answer checks.

Now, mass needed.
moles NaCNO = (0.711 mol/L)(0.1000 L) = 0.0711 moles
molar mass NaCNO = 65.01 g/mol
mass NaCNO = (0.0711 mole)(65.01 g)/(1 mole) = 4.622 g NaCNO

13.

We can use the Henderson-Hasselbalch equation to solve this problem.
pH = pKa + log{[CH3CH2COO^-]/[ CH3CH2COOH]}
pKa [CH3CH2COOH] = 4.88

therefore;

Moles of CH3CH2COONa or CH3CH2COO- = 0.050*0.12

= 0.006 moles

And CH3CH2COOH = 200/1000 *0.04

= 0.008

then

pH = pKa + log{[ 0.006] /[0.008]}

= 4.88 + log 0.75

= 4.88- 0.125

= 4.755