Asked by Jude
A buffer was made by mixing 0.1522 moles of HCO2H with 0.1832 moles LiHCO2- and diluting to exactly 1L. What will be the pH after addition of 10ml of 0.3657M NaOH to 50mL of the buffer. Ka(HCO2H) = 1.8*10-4
Answers
Answered by
DrBob222
In 50 mL of the buffer you will have
millimoles COO- = 50 mL x 0.18322M = 9.160
mmoles HCOOH = 50mL x 0.1522 = 7.610
mmols NaOH added = 10 mL x 0.3657 = 3.657
...........HCOOH + OH^= ==> HCOO^- + H2O
initial....7.610....0........9.160
add................3.657............
change.....-3.657..-3.657..+3.657..3.657
equil..... sum.......0......sum......sum
Substitute into the HH equation and solve for pH.
millimoles COO- = 50 mL x 0.18322M = 9.160
mmoles HCOOH = 50mL x 0.1522 = 7.610
mmols NaOH added = 10 mL x 0.3657 = 3.657
...........HCOOH + OH^= ==> HCOO^- + H2O
initial....7.610....0........9.160
add................3.657............
change.....-3.657..-3.657..+3.657..3.657
equil..... sum.......0......sum......sum
Substitute into the HH equation and solve for pH.
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