Question

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A.What is the pH of 0.010 M of Na2CO3 solution? (Ka1 and Ka2 of H2CO3= 4.3...

A.What is the pH of 0.010 M of Na2CO3 solution? (Ka1 and Ka2 of H2CO3= 4.3 x 10-7 and 4.8 x 10^-11).

B.Calculate the values of pH when 40.0 mL of 0.0250M benzoic acid (HC7H5O2, Ka=6.3x10-5) is titrated with 10.0 mL of 0.050 M NaOH solution?

Expert Solution

A)
CO3-2 + H2O -------------------> HCO3- + OH-

0.01 0 0

0.01-x x x

Kb1 = Kw / Ka2 = 10^-14 / 4.8 x 10^-11 = 2.08 x 10^-4

2.08 x 10^-4 = x^2 / 0.01 -x

x^2 + 2.08 x 10^-4 x - 2.08 x 10^-6 = 0

x = 1.34 x 10^-3

x = [OH-] = 1.34 x 10^-3 M

pOH = -log (1.34 x 10^-3)

pOH = 2.68

pH + pOH = 14

pH = 11.32

B)

millimoles of acid = 40 x 0.0250 = 1

millimoles of NaoH = 10 x 0.05 = 0.5

it is half equivalence point .

here pH = pKa

pKa = -log Ka

pKa = 4.20

pH = 4.20

queen_honey_blossom answered 2 years ago

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