For carbonic acid (H₂CO₃), K_a1 = 4.30 × 10^-7 and K_a2 = 5.62 × 10^-11. Calculate the pH of a 0.50 M solution of Na₂CO₃.

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The correct answer is A because the pH of a 0.50 M solution of Na2CO3 can be calculated using the following steps:1. Write the dissociation equations for carbonic acid:H2CO3(aq) + H2O(l) <=> H3O+(aq) + HCO3-(aq) (Ka1 = 4.30 × 10^-7)HCO3-(aq) + H2O(l) <=> H3O+(aq) + CO32-(aq) (Ka2 = 5.62 × 10^-11)2. Since Na2CO3 is a salt of a weak acid (H2CO3) and a strong base (NaOH), it will undergo hydrolysis in water to produce OH- ions. The reaction can be represented as:CO32-(aq) + H2O(l) <=> HCO3-(aq) + OH-(aq)3. The Kb for the reaction in step 2 can be calculated using the relationship Kb = Kw/Ka, where Kw is the ion product constant for water (1.0 × 10^-14) and Ka is the acid dissociation constant for H2CO3 (4.30 × 10^-7). Therefore, Kb = 2.32 × 10^-8.4. The pH of the solution can be calculated using the following equation:pH = 14 - pOHpOH = -log[OH-]5. The concentration of OH- ions can be calculated using the Kb expression:Kb = [HCO3-][OH-]/[CO32-][OH-] = Kb[CO32-]/[HCO3-]6. The concentration of HCO3- ions can be calculated using the dissociation equation for H2CO3:Ka1 = [H3O+][HCO3-]/[H2CO3][HCO3-] = Ka1[H2CO3]/[H3O+]7. The concentration of H2CO3 can be calculated using the ICE table:```Initial: 0.50 M 0 M 0 MChange: -x +x +xEquilibrium: (0.50-x) M x M x M```8. Substituting the expressions for [HCO3-] and [H2CO3] into the Kb expression, we get:Kb = Kb(0.50-x)/xx = 4.64 × 10^-9 M9. Substituting the value of x into the expression for [OH-], we get:[OH-] = 2.32 × 10^-8(0.50 M)/(4.64 × 10^-9 M)[OH-] = 2.50 × 10^-6 M10. Finally, substituting the value of [OH-] into the equation for pOH, we get:pOH = -log(2.50 × 10^-6)pOH = 5.6011. Therefore, the pH of the solution is:pH = 14 - pOHpH = 14 - 5.60pH = 8.40

For Carbonic Acid (H2CO3), Ka1 = 4

For Carbonic Acid (H₂CO₃), K_a1 = 4