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Determine the ph of a 0.18 m h2co3 solution. carbonic acid is a diprotic acid whose ka1 = 4.3 × 10-7 and ka2 = 5.6 × 10-11. determine the ph of a 0.18 m h2co3 solution. carbonic acid is a diprotic acid whose ka1 = 4.3 × 10-7 and ka2 = 5.6 × 10-11. 4.31 5.50 11.00 10.44 3.56

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Dejanras
Answer is: ph value is 3.56.
Chemical reaction 1: H₂CO₃(aq) ⇄ HCO₃⁻(aq) + H⁺(aq); Ka₁ = 4,3·10⁻⁷.
Chemical reaction 2: HCO₃⁻(aq) ⇄ CO₃²⁻(aq) + H⁺(aq); Ka₂ = 5,6·10⁻¹¹.
c(H₂CO₃) = 0,18 M.
[HCO₃⁻] = [H⁺] = x.
[H₂CO₃] = 0,18 M - x.
Ka₁ = [HCO₃⁻] · [H⁺] / [H₂CO₃].
4,3·10⁻⁷ = x² / (0,18 M -x).
Solve quadratic equation: x = [H⁺] =0,000293 M.
pH = -log[H⁺] = -log(0,000293 M).
pH = 3,56; second Ka do not contributes pH value a lot.


mickymike92

The pH of the 0.18 M H2CO3 solution is 3.55

Acids dissociation constant, Ka

  • The acid dissociation constant (Ka) is a quantitative measure of the strength of an acid in solution.

It is a ratio of the products in an acid dissociation to the reactant.

Ka of a diprotic acid

For a diprotic acid, whose dissociation produces two H+ ions, there are two Ka values.

The dissociation of H2CO3 is given below:

  • H2CO3 ----> HCO3- + H+ ; Ka1 = 4.3 × 10^-7
  • HCO3- ----> CO3^2- + H+; Ka2 = 5.6 × 10^-11

Since Ka2 <<< Ka2, the second ionization is negligible.

The pH of the solution depends on Ka1.

Using an ICE table

H2CO3 ------> HCO3- + H+

Initial concentration of H2CO3 = 0.18 M

Initial concentration of HCO3- = 0

Initial concentration of H+ = 0

Change in concentration of H2CO3 = -X

Change in concentration of HCO3- = +X

Change in concentration of H+ = +X

Equilibrium concentration of H2CO3 = 0.18 - X

Equilibrium concentration of HCO3- = X

Equilibrium concentration of H+ = X

Assuming X is very small, 0.18 - X = 0.18

Ka1 = [HCO3-] [H+]/ H2CO3

4.37 × 10^-7 = X^2 / 0.18

Solving for X

X = 2.80

Therefore, [H+] = 2.80 × 10^-4 M

pH = - log 2.80 × 10^-4 M

pH = 3.55

Therefore, the pH of the 0.18 M H2CO3 solution is 3.55

Learn more about pH and Ka at: https://brainly.com/question/9465562

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