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'm not sure how to go about solving this problem.
Calculate the pH of a 0.15 M solution of Na2CO3.
I'm not sure how to go about solving this problem.
HPV
Favorite Answer
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.
Gipsy
This Site Might Help You.
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'm not sure how to go about solving this problem.
?
Ka Of Carbonic Acid
Mila
For the best answers, search on this site https://shorturl.im/RTCNl
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+