Question

In: Chemistry

for the diuretic weak acid H2A Ka1=2.2x10^-6 and Ka2=8.6x10^-9. what is the pH of a .0500M...

for the diuretic weak acid H2A Ka1=2.2x10^-6 and Ka2=8.6x10^-9. what is the pH of a .0500M solution of H2A? what are the equilibrium concentrations of H2A and A^2- in this solution?

Expert Solution

[H2A] = 0.0500M Ka1= 2.2x10^-6 Ka2 = 8.6x10^-9

H2A ---------------- HA- + H+

Initial 0.0500 0 0

change -x +x +x

equilibrum 0.0500-x +x +x

Ka1 = [HA-][H+]/[H2A]

2.2x10^-6 = x*x/( 0.0500-x)

for solving the equation

x= 0.00033

[HA-] = 0.00033M

[H+] = 0.00033 M

-log[H+] = -log(0.00033)

PH= 3.48

[H2A] = 0.0500 - 0.00033 = 0.04967M

[H2A] = 0.0497M

HA- ---------------- H+ + A-2

0.00033 0.00033 x

Ka2 = [H+][A-2]/[HA-]

8.6x10^-9 = 0.00033x [A-2]/0.00033

[A-2] = 8.6x10^-9M

[H2A] = 0.0497M

PH = 3.48

queen_honey_blossom answered 2 years ago

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