Question

Calculate the change in pH when 4.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq). Delta pH= ?

Calculate the change in pH when 4.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution. Delta pH= ?

Answer 1

i) pH of original solution

At the equal concentration of weak acid and its conjucate base pH of the buffer solution would be equal to pKa of the weak acid

pKa of NH4Cl = 9.25

Therefore

pH of original solution = 9.25

ii) ∆pH for the addition of HCl

Initial moles of NH4⁺ = ( 0.100mol/1000ml) × 100ml = 0.0100mol

Initial moles of NH3 = (0.100mol/1000ml) × 100ml = 0.0100mol

moles of HCl added = (0.100mol/1000ml) ×4ml = 0.0004mol

HCl reacts with NH3

HCl(aq) + NH3(aq) ---------> NH4⁺(aq) + Cl^-(aq)

0.0004moles of HCl reacts with 0.0004moles of NH3 to give 0.0004 moles of NH3

After addition of HCl

number of moles of NH4⁺ = 0.0100mol + 0.0004mol = 0.0104mol

number of moles of NH3 = 0.0100mol - 0.0004mol = 0.0096mol

[NH4⁺] = (0.0104mol/100ml)× 1000ml = 0.1040M

[NH3] = (0.0096mol/100ml) × 1000ml = 0.0960M

Henderson - Hasselbalch equation

pH = pKa + log([A^-]/[HA])

pH = pKa + log([NH3]/[NH4⁺])

pH = 9.25 + log(0.0960M/0.1040M)

pH = 9.25 - 0.03

Therefore

∆pH = - 0.03

iii) ∆pH for the addition of NaOH

Moles of NaOH added =.(0.100!mol/1000mol) × 4.00mol = 0.0004mol

NaOH reacts with NH4⁺

NaOH(aq) + NH4⁺(aq) --------> NH3(aq) + Na⁺(aq)

After addition of NaOH

number of moles of NH4⁺ = 0.0100 - 0.0004 = 0.0096mol

number of moles of NH3 = 0.0100 + 0.0004 = 0.0104mol

[NH4⁺] = (0.0096mol/100ml) ×1000ml = 0.0960M

[NH3] = (0.0104mol/100ml) ×1000ml = 0.1040M

Applying Henderson - Hasselbalch equation

pH = pKa + log([A^-] /[ NH3])

pH = pKa + log([NH3]/[NH4⁺])

pH = 9.25 + log( 0.1040M/0.096M)

pH = 9.25 + 0.03

Therefore

∆pH = + 0.03