1)
Ka = 3.04*10^-7
pKa = - log (Ka)
= - log(3.04*10^-7)
= 6.517
use:
pH = pKa + log {[conjugate base]/[acid]}
= 6.517+ log {0.796/0.311}
= 6.925
Answer: 6.925
2)
mol of KOH added = 0.158M *0.318 L = 0.05024 mol
HA will react with OH- to form A-
Before Reaction:
mol of A- = 0.796 M *1.0 L
mol of A- = 0.796 mol
mol of HA = 0.311 M *1.0 L
mol of HA = 0.311 mol
after reaction,
mol of A- = mol present initially + mol added
mol of A- = (0.796 + 0.05024) mol
mol of A- = 0.8462 mol
mol of HA = mol present initially - mol added
mol of HA = (0.311 - 0.05024) mol
mol of HA = 0.2608 mol
Ka = 3.04*10^-7
pKa = - log (Ka)
= - log(3.04*10^-7)
= 6.517
since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 6.517+ log {0.8462/0.2608}
= 7.028
Answer: 7.028
3)
mol of KOH added = 0.158M *0.636 L = 0.1005 mol
HA will react with OH- to form A-
Before Reaction:
mol of A- = 0.796 M *1.0 L
mol of A- = 0.796 mol
mol of HA = 0.311 M *1.0 L
mol of HA = 0.311 mol
after reaction,
mol of A- = mol present initially + mol added
mol of A- = (0.796 + 0.1005) mol
mol of A- = 0.8965 mol
mol of HA = mol present initially - mol added
mol of HA = (0.311 - 0.1005) mol
mol of HA = 0.2105 mol
Ka = 3.04*10^-7
pKa = - log (Ka)
= - log(3.04*10^-7)
= 6.517
since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 6.517+ log {0.8965/0.2105}
= 7.146
Answer: 7.146
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