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2 years ago

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

Chemistry

1 answer:

fgiga [73]2 years ago

5 0

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

Learn more about concentration here-

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The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

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$O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol$ ..[1]

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2 × [4] = [2]- (3 ) × [1] - (2) × [3]

$2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}$

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