A 1.0-L buffer solution contains 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.
Calculate the pH of the solution, upon addition of 0.070 mol of NaOH to the original buffer.
Express your answer using two decimal places.
mol of NaOH added = 0.07 mmol
HC2H3O2 will react with OH- to form C2H3O2-
Before Reaction:
mol of C2H3O2- = 0.1 mmol
mol of HC2H3O2 = 0.1 mmol
after reaction,
mol of C2H3O2- = mol present initially + mol added
mol of C2H3O2- = (0.1 + 0.07) mmol
mol of C2H3O2- = 0.17 mmol
mol of HC2H3O2 = mol present initially - mol added
mol of HC2H3O2 = (0.1 - 0.07) mmol
mol of HC2H3O2 = 0.03 mmol
Ka = 1.8*10^-5
pKa = - log (Ka)
= - log(1.8*10^-5)
= 4.745
since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.745+ log {0.17/3*10^-2}
= 5.498
Answer: 5.50
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