Calculate the pH of a 0.20 M H₂SO₃ solution that has the stepwise dissociation constants K_a1 = 1.5 × 10^-2 and K_a2 = 6.3 × 10^-8. A)1.26 B)1.32 C)1.82 D)2.52

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First, we can write out the chemical equation for the dissociation of H₂SOH as follows: H₂SOH ⇌ H+ + ₂SO- The equilibrium constant for this reaction is K_a1 = [H+][₂SO-]/[H₂SOH], which we can rearrange to obtain an expression for [H+] in terms of the concentration of H₂SOH: [H+] = K_a1[H₂SOH]/[₂SO-] At an initial concentration of 0.20 M, the concentration of H₂SOH is also 0.20 M, so we can substitute these values into the equation above to obtain: [H+] = (1.5 × 10^-2)(0.20 M)/(0.20 M) = 1.5 × 10^-2 Next, we need to consider the dissociation of ₂SO-. Writing out the chemical equation as before, we have: ₂SO- ⇌ H+ + ₃SO2- The equilibrium constant for this reaction is K_a2 = [H+][₃SO2-]/[₂SO-], which we can rearrange to obtain an expression for [H+] in terms of the concentration of ₂SO-: [H+] = K_a2[₂SO-]/[₃SO2-] Using the law of mass action, we know that the concentration of ₃SO2- is equal to the concentration of H+ that was produced in the first step (since one mole of H+ is generated for every mole of ₂SO- that reacts), so we can substitute 1.5 × 10^-2 for [H+] in the equation above to obtain: [H+] = (6.3 × 10^-8)(0.20 M)/(1.5 × 10^-2) = 1.32 Therefore, the pH of the solution is 1.32.

Calculate the PH of a 0