Question

calculate the ph of a 0.020 m carbonic acid solution, h2co3(aq), that has the stepwise dissociation constants ka1

Answer 1

The calculated pH is 3.79. therefore, the solution is acidic.

No, carbonic acid is not a strong acid. H2CO3 is a weak acid that dissociates into a proton (H+ cation) and a bicarbonate ion (HCO3- anion). This compound only partly dissociates in aqueous solutions.

H2 CO3    =    H (+) + HCO3(-)          Ka1 = 4.3 * 10^ -7

   0.06 - x              x          x

Ka1 = x^2 / (0.06 - x) = 4.3 * 10^ - 7

A low Ka => x << 0.06 => 0.06 -x ≈ 0.06

=> Ka1 ≈ x^2 / 0.06 => x^2 ≈ 0.06 * Ka1 = 0.06 * 4.3 * 10^-7

=> x ≈ √ [ 2.58 * 10 ^ -8] = 1.606 * 10^ - 4 = 0.0001606

Second dissociation

HCO3(-)    =   H (+) + CO3(2-)         Ka2 = 5.6 * 10^ - 11

0.0001606 - y            y             y

Ka2 ≈ y^2 / 0.0001606 => y = √ [0.0001606 * 5.6* 10^ -11]

y = 9.48 * 10^ -8

An acidic solution has a high concentration of hydrogen ions (H +start superscript, plus, end superscript), greater than that of pure water.

[H+] = x + y = 1.607 * 10^ -4

pH = - log [H+] = 3.79

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