Question

Calculate the pH of the 0.250 M NH3/0.5 M NH4Cl buffer system. What is the pH after the addition of 2.0 mL of 0.250 M NaOH to 18.0 mL of the buffer solution? After adding 10 more mL of 0.25 M NaOH what is the pH?

Answer 1

pH of buffer solution of 0.25 M NH3 and 0.5 molar NH4Cl

we know Kb of NH3 =  1.8 x 10^-5

NH3 + H2O <------> NH4+ + OH-

NH3 + H+ -> NH4+

NH4+ + OH- -> NH3

The pH of a buffer can be calculated using the Henderson-Hasselbalch equation. Here's the form of the equation for a base and its conjugate acid:

pOH = pKb + log [NH4+] / [NH3}

pOH = 4.74 + log [0.5 / 0.25] = 5.04

pH = 14 - 5.04 = 8.95

The NaOH that's added reacts the NH4+, converting it to NH3.

We can find the new concentration of NH4+ by subtracting the amount that reacted:

Moles NH3 = 0.018 L x 0.25 M = 0.0045 moles

Moles NH4+ = 0.018 L x 0.5 M = 0.009 moles

Moles OH- = 0.002 L x 0.25 M = 0.0005 moles

NH4+ + OH- >> NH3 + H2O

moles NH3 = 0.0045 + 0.0005 moles = 0.005

moles NH4+ = 0.009 - 0.0005 = 0.0085

total volume = 2 + 18 = 20 mL = 0.02 L

concentration NH3 = ( 0.005 / 0.02 L) = 0.25 M

concentration NH4+ = (0.0085 / 0.02 L) = 0.425 M

pOH = 4.74 + log [0.425 / 0.25] = 4.97

pH = 14 - 4.97 = 9.03

same for 10 ml of 0.25 M NaOH

Moles OH- = 0.01 L x 0.25 M = 0.0025 moles + (Moles OH- = 0.002 L x 0.25 M = 0.0005 moles ) = 0.003 moles

moles NH3 = 0.0045 + 0.0025 moles +0.0005 moles = 0.0075

moles NH4+ = 0.009 - (0.0025 + 0.0005) = 0.006

total volume = 10 + 2 + 18 = 30 mL = 0.03 L

concentration NH3 = ( 0.0075 / 0.03 L) = 0.25 M

concentration NH4+ = (0.006 / 0.03 L) = 0.2 M

pOH = 4.74 + log [0.2 / 0.25] = 4.64

pH = 14 - 4.64 = 9.35