1)
mol of HI added = 0.0568 mol
NH3 will react with H+ to form NH4+
Before Reaction:
mol of NH3 = 0.457 M *0.25 L
mol of NH3 = 0.1143 mol
mol of NH4+ = 0.229 M *0.25 L
mol of NH4+ = 0.0573 mol
after reaction,
mol of NH3 = mol present initially - mol added
mol of NH3 = (0.1143 - 0.0568) mol
mol of NH3 = 0.0575 mol
mol of NH4+ = mol present initially + mol added
mol of NH4+ = (0.0573 + 0.0568) mol
mol of NH4+ = 0.1141 mol
since volume is both in numerator and denominator, we can use mol
instead of concentration
Kb = 1.8*10^-5
pKb = - log (Kb)
= - log(1.8*10^-5)
= 4.745
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 4.745+ log {0.1141/5.745*10^-2}
= 5.043
use:
PH = 14 - pOH
= 14 - 5.0425
= 8.9575
Answer: 8.96
2)
mol of KOH added = 0.0313 mol
NH4+ will react with OH- to form NH3
Before Reaction:
mol of NH3 = 0.443 M *0.225 L
mol of NH3 = 0.0997 mol
mol of NH4+ = 0.314 M *0.225 L
mol of NH4+ = 0.0707 mol
after reaction,
mol of NH3 = mol present initially + mol added
mol of NH3 = (0.0997 + 0.0313) mol
mol of NH3 = 0.131 mol
mol of NH4+ = mol present initially - mol added
mol of NH4+ = (0.0707 - 0.0313) mol
mol of NH4+ = 0.0394 mol
since volume is both in numerator and denominator, we can use mol
instead of concentration
Kb = 1.8*10^-5
pKb = - log (Kb)
= - log(1.8*10^-5)
= 4.745
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 4.745+ log {3.935*10^-2/0.131}
= 4.222
use:
PH = 14 - pOH
= 14 - 4.2225
= 9.7775
Answer: 9.78
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