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What is the pH of a 0.1M solution of monosodium succinate pk1= 4.19 pk2= 5.56

What is the pH of a 0.1M solution of monosodium succinate
pk1= 4.19
pk2= 5.56

Solutions

Expert Solution

Monosodium succinate = C4H5NaO4

    C4H5NaO4 -----------------------C4H5O4- + Na+

               C4H5O4-     H2O ----------------- C4H5O4H + OH-

                  0.1                                              0                  0

                   -x                                               +x               +x

                 0.1-x                                          +x                   +x

Ka = [C4H5OH][OH-]/[C4H5O4-]

Pka1 = 4.19

-log(Ka1) = 4.19

Ka1 = 10^-4.19

Ka1= 6.46x10^-5

6.46x10^-5 = x*x/(0.1-x)

for solving the equation

x= 0.0025

[OH-] =0..0025M

-log[OH-] = -log(0.0025)

POH= 2.60

PH+POH=14

PH= 14 - POH

PH = 14- 2.60

PH = 11.4

queen_honey_blossom answered 2 years ago
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