500.0 mL of 0.110 M NaOH is added to 615 mL of 0.200 M weak acid...

500.0 mL of 0.110 M NaOH is added to 615 mL of 0.200 M weak acid (Ka = 7.63

Expert Solution

millimoles of acid = 0.2 x 615 =123

millimoles of base = 500 x 0.11=55

HA + NaOH ---------------------------> NaA + H2O

123 55 0 0 -----------------------------before reaction

68 0 55 55 ------------------------after reaction

in this mixture salt NaA and acid HA remains . it can form buffer

pKa = -log Ka = -log(7.63

queen_honey_blossom answered 2 months ago

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