Question

1.Calculate the pH of a solution obtained by mixing 100 ml of 1 M HEPES and 25 ml of 1 M NaOH and adjusting the volume to one liter with water.

2. Calculate the pH of a solution obtained by mixing 100 ml of 1 M NH3 (pKa 9.25) and 60 ml of 1 M HCl and adjusting the volume to one liter with water. (Answer 9.07)

I want to know the specific calculation to get a final answer. Thank you.

Answer 1

a)When you add NaOH to the CH3COOH solution you form CH3COONa. You then have a solution of a weak acid and a salt of this acid. This is a buffer solution . The pH of the buffer solution is calculated using the Henderson - Hasselbalch equation.
pH = pKa + log ([salt] /[acid])
Some preliminary calculations :
pKa = -log Ka = - log (1.7*10^-5) = 4.77

Mol CH3COOH in 100mL of 0.1M solution = 100/1000*0.1 = 0.01 mol
Mol NaOH in 25mL of 0.1M solution = 25/1000*0.1 = 0.0025 mol
The NaOH reacts with the CH3COOH in 1:1 ratio
Therefore you produce 0.0025 mol CH3COONa
and remaining unreacted is 0.0075 mol CH3COOH
In total volume 125mL = 0.125 L
Molarity of CH3COONa = 0.0025/0.125 = 0.02M
Molarity of CH3COOH = 0.0075 / 0.125 = 0.06M
Now use the H-H equation:
pH = pKa + log ([salt] /[acid])
pH = 4.77 + log ( (0.02/0.06)
pH = 4.77 + log 0.333
pH = 4.77 + (-0.48)
pH = 4.29

b)

moles NH3 = 0.100 L x 0.300 M = 0.0300
moles HCl = 0.06 L x 0.100 M = 0.006

NH3 + +H+ >> NH4+
Moles NH3 = 0.0300 - 0.0100 = 0.024
moles NH4+ = 0.0100
total volume = 0.24 l
cocnentration NH3 = 0.0200 / 0.200 = 0.100 M
concentration NH4+ = 0.0100 / 0.200 = 0.0500 M

pKa = 9.25

pOH = 9.25 + log 0.0500 / 0.100 = 9.07