For carbonic acid (H2CO3) the ka1 = 4.30 x 10^-7 and the Ka2 = 5.62 x 10^-11.?

The Ka1 reaction is H2CO3 losing a proton: H2CO3 + H2O <==> H3O+ + HCO3-

The Ka2 reaction is HCO3- losing a proton: HCO3- + H2O <==> H3O+ + CO3 2-

Since your solution contains only Na2CO3, we need to use the Ka2 data. CO3 2- is the conjugate base of HCO3- so this is actually a Kb problem.

Ka HCO3- = 5.62 x 10^-11

Kb CO3 2- = Kw / Ka = (1 x 10^-14 / 5.62 x 10^-11) = 1.78 x 10^-4

CO3 2- will act as a base (accept H+ ion) in H2O to form small amounts of HCO3- and OH-.

Molarity . . . . . . .CO3 2- + H2O <==> HCO3- + OH-

Initial . . . . . . . . . .0.15 . . . . . . . . . . . . . . .0 . . . . . .0

Change . . . . . . . . .-x . . . . . . . . . . . . . . . .x . . . . . .x

At Equilibrium . . .0.15 - x . . . . . . . . . . . . .x . . . . . .x

Kb = [HCO3-][OH-] / [CO3 2-] = (x)(x) / (0.15 - x) = 1.78 x 10^-4

Because Kb is small, the -x term after 0.15 can be dropped to simplify the math.

x^2 / 0.15 = 1.78 x 10^-4

x^2 = 2.67 x 10^-5

x = 5.17 x 10^-3 = [OH-]

pOH = -log [OH-] = -log (5.17 x 10^-3) = 2.287

pH + pOH = 14

pH = 14 - pOH = 14 - 2.287 = 11.71 . . .definitely alkaline.

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RE:

For carbonic acid (H2CO3) the ka1 = 4.30 x 10^-7 and the Ka2 = 5.62 x 10^-11.?

Calculate the pH of a 0.15 M solution of Na2CO3.

I&#39;m not sure how to go about solving this problem.

Ka Of Carbonic Acid

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Ka1 = 4.3x10^-7 = [H+][HCO3-]/[H2CO3] 4.3x10^-7 = x^2 / 5.45x10^-4 - x x^2 + 4.3x10^-7x - 2.34x10^-10 = 0 x = [H+] = [HCO3-] = 1.51x10^-5M Ka2 = 4.8x10^-11 = [H+][CO3 2-]/[HCO3-] because the Ka is so small, we can ignore x in the denominator. AND, the Ka2 is so small that the added [H+] will be negligible so you could ignore it all together. your choice but not always the best one. the equation becomes: 4.8x10^-11 = x^2 / 1.51x10^-5 x = [H+] = 2.68x10^-8M assuming that the volume = 1L, 1.51x10^-5moles + 2.68x10^-8moles / 1L = 1.51x10^-5M H+