a. pH of weak acid = 1/2(pka-logC) pka1 = -log(7.5*10^-3) = 2.12 = 1/2(2.12-log0
ID: 699512 • Letter: A
Question
a. pH of weak acid = 1/2(pka-logC)
pka1 = -log(7.5*10^-3) = 2.12
= 1/2(2.12-log0.1)
pH = 1.56
b. no of mol of H3PO4 = 30*0.1 = 3 mmol
no of mol of KOH = 15*0.2 = 3 mmol
concentration of salt(NaH2PO4) formed = 3/45 = 0.067 M
pka1 = 2.12
pH = 7+1/2(pka1+logC)
= 7+1/2(2.12+log0.067)
= 7.47
c. no of mol of H3PO4 = 30*0.1 = 3 mmol
no of mol of KOH = 42*0.2 = 8.4 mmol
H3PO4+NaOH ---> NaH2PO4+H2O ( 3 mmol NaOH consumed)
NaH2PO4+NaOH ---> Na2HPO4 + H2O ( 3 mmol NaOH formed , 2.4 mmol consumed)
Na2HPO4+ NaOH ---> Na3PO4 + H2O ( 0mly 2.4 mmol formed)
so that,
No of mol of Na2HPO4 present finally = 0.6 mmol
No of mol of Na3PO4 present finally = 2.4 mmol
pka3 = -log(4.8*10^-13) = 12.32
pH = pka3+log(Na3PO4/Na2HPO4)
= 12.32+log(2.4/0.6)
= 12.92
Explanation / Answer
Page 11 of 15 11. Phosphoric acid is a triprotic acid with the chemical formula H,PO4 The acid dissociation constants for phosphoric acid are as follows: Kal-7.5 x 10-3 ; Ka2-6.2 x 10"";Ka=4.8 x 10"3 If you are given 30.00 mL of a 0.100 M solution of phosphoric acicd: a. What is the pH of the solution? (6 points) b. What is the pH of the resulting solution when 15.00 mL of 0.200 M KOH is added to 30.00 mL of 0.100 M phosphoric acid? (6 points) c. What is the pH of the resulting solution when 42.00 mL of 0.200 M KOH is added to the original 30.00 mL solution of 0.100 M phosphoric acid? 6 points)
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