Question

Calculate the molar concentration of Ag and SO, at the equilibrium. [10 marks] iv) The sulfate anion (SO42) is a very weak base undergoing the following (balanced) acid-base equilibrium reaction in water (KB = 8.3x10 "); SO42- + H2O = HSO4 + OH Calculate the pH of the Ag2SO4 solution (shown above) at the equilibrium (hint: disregard the autodissociation of water). [20 marks) Justify your answers by including workings, calculations and units.

Answer 1

∴ assuming that C SO42- = 1.00 M:

• Ag2SO4 solution, pH = 7.96
• on equilibrium:

• [ Ag+ ] = 2 * C SO42- = 2.00 M
• [ SO42- ] = C SO42- - [ OH- ] = 1.00 M

Step-by-step explanation:

• Ag2SO4 → 2Ag+ + SO42-

C 2*C C .................. equilibirium

• SO42- + H2O ↔ HSO4- + OH-

∴ Kb = ( [ OH- ] * [ HSO4- ] ) / [ SO42- ] = 8.3 E-13,,,,most important equilibrium

• HSO4- + H2O ↔ H2SO4 + OH-

mass balance:

⇒ C SO42- = [ HSO4- ] + [ SO42- ]......(1)

charge balance:

⇒ [ Ag+ ] = [ HSO4- ] + 2[SO42-] + [ OH- ]

∴ [ Ag+ ] = 2*C SO42-

⇒ 2 C SO42- = [ HSO4- ] + 2[SO42-] + [ OH- ]...........(2)

∴ ((1) * 2) - (2):

⇒ 0 = [ HSO4- ] - [ OH- ]

⇒ [ HSO4- ] = [ OH- ]........(3)

(3) in Kb:

⇒ 8.3 E-13 = [ OH- ]² / [ SO42- ] = [ OH- ]² / ( C SO42- - [ OH- ] )

⇒ [ OH- ]² + 8.3 E-13 [OH- ] - 8.3 E-13*C SO42- = 0

assuming C SO42- = 1.00 M:

⇒ [ OH- ] = 9.11 E-12 M

⇒ pOH = 6.04

⇒ pH = 7.96

on equilibrium:

• [ Ag+ ] = 2 * C SO42- = 2.00 M
• [ SO42- ] = C SO42- - [ OH- ] = 1.00 M