Question

a. A buffer solution contains 0.274 M NH₄Br and 0.379 M NH₃ (ammonia). Determine the pH change when 0.103 mol HBr is added to 1.00 L of the buffer.
pH after addition − pH before addition = pH change =

b. A buffer solution contains 0.432 M NaH₂PO₄ and 0.209 M K₂HPO₄. Determine the pH change when 0.113 mol KOH is added to 1.00 L of the buffer.
pH after addition − pH before addition = pH change =

Answer 1

a)

i) pH before addition of HBr

Henderson - Hassrlbalch equation is

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

A^- = conjucate base , NH3

HA = weak acid , NH4⁺

pKa = pKa of weak acid , 9.26

substituting the values

pH = 9.26 + log(0.379M/ 0.274M)

pH = 9.26 + 0.14

pH = 9.40

ii) pH after addition of HBr

Initial moles of NH3 = 0.379mol

Initial moles of NH4⁺ = 0.274mol

moles of HBr added = 0.103mol

HBr as an acid react with base NH3

NH3 + H⁺ --------> NH4⁺

After addition of HBr

moles of NH3 = 0.379mol - 0.103mol = 0.276mol

moles of NH4⁺ = 0.274mol + 0.103mol = 0.377mol

[NH3] = 0.276M

[NH4⁺] = 0.377M

Applying Henderson- Hasselbalch equation

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

pH = 9.26 + log(0.276M/ 0.377M)

pH = 9.26 - 0.14

pH = 9.12

iii) pH change

pH change = 9.12 - 9.40

pH change = - 0.28

b)

i) before addition of KOH

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

pH = 7.21 + log( 0.209M/ 0.432)

pH = 6.89

ii) pH after addition of NaOH

Initial moles of H2PO4^- = 0.432 mol

Initial moles of HPO4^2- = 0.209 mol

moles of KOH added = 0.113mol

KOH as a base react with acid H2PO4^-

OH^- + H2PO4^- --------> HPO4^- + H2O

Aftet adfition NaOH

moles of H2PO4^- = 0.432mol - 0.113 mol = 0.319mol

moles of HPO4^- = 0.209mol + 0.113mol = 0.322mol

[H2PO4^-] =0.319M

[HPO4^2-​​​​​​] = 0.322M

Applying Henderson -Hasselbach equation

pH = 7.21 + log( 0.322M / 0.319M)

pH = 7.21 + 0.004

pH = 7.21

iii) pH change

pH change = 7.21 - 6.89

pH change = 0.32