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In: Chemistry

Consider the titration of 100.00mL of 1.00 M H3PO4 with 1.00M NaOH. ka1= 7.5x10^-3, ka2= 6.2x10^-8,...

Consider the titration of 100.00mL of 1.00 M H3PO4 with 1.00M NaOH. ka1= 7.5x10^-3, ka2= 6.2x10^-8, ka3= 4.2x10^-13 a. calculate the pH before any base is added. Hint: first write down the major species and pick out the strongest acid. b. calculate the pH after 25.00 mL of base is added C. calculate pH after 50.00mL base is added d. calculate pH after 100.0 mL base is added e. calculate pH after 150.00 mL base is added f. calculate pH after 200 mL base is added g. calculate pH after 225.00 mL base is added h. calculate pH after 250.00 ml base is added i. calcualte pH after 300.00 mL base is added j. calculate pH after 350.00 mL base is added please show work!!! so I can understand it

Solutions

Expert Solution

pKa1 = -log Ka1 = -log (7.5 x 10^-3)

pKa1 = 2.12

pKa2 = 7.21

pKa3 = 12.38

a. calculate the pH before any base is added

H3PO4 ----------------------------> H2PO4- + H+

1.0                                           0            0 -------------------> initial

-x                                           + x           +x ---------------> change

1.0-x                                       x             x -----------------> equilibrium

Ka1 = x^2 / 1.0 -x = 7.5 x 10^-3

by solving this   x = 8.3 x 10^-2

[H+] = x = 8.3 x 10^-2 M

pH   = -log [H+] = -log (8.3 x 10^-2 )

pH = 1.08

b. calculate the pH after 25.00 mL of base is added

millimoles of H3PO4 = 100 x 1 = 100

millimoles of NaOH = 25 x 1 = 25

H3PO4 + NaOH -----------------------------> NaH2Po4 + H2O

100             25                                           0               0

75              0                                             25

pH = pKa1 + log [NaH2PO4 / H3PO4]

pH = 2.12 + log (25 / 75)

pH = 1.64

C. calculate pH after 50.00mL base is added

it is first half equivalence point : here pH = pKa1

pH = 2.12

d. calculate pH after 100.0 mL base is added

it is first equivalence point

here

pH = 1/2 [pKa1 + pKa2 ]

pH = 1/2 [2.12+ 7.21]

pH = 4.67

e. calculate pH after 150.00 mL base is added

it is second half equivalence here pH = pKa2

pH =7.21

f. calculate pH after 200 mL base is added

it is second equivalence point

here pH = 1/2 [pKa2 + pKa3]

       pH = 1/2 [7.21 + 12.38]

      pH = 9.80

queen_honey_blossom answered 2 years ago
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