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.

∴ assuming that C SO42- = 1.00 M:

Explanation:

          C            2*C          C equilibirium

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

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:

i believe it is 40–40–20%

explanation: