Calculate the percent ionization of formic acid (HCO₂H)in a solution that is 0.322 M in formic acid and 0.178 M in sodium formate (NaHCO₂). The K_a of formic acid is 1.77 × 10^-4. A)35.6 B)0.1011 C)10.8 D)1.03 × 10^-3 E)3.488

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First, we need to write the balanced equation for the ionization of formic acid: HCO₂H ⇌ H+ + CO₂- Next, we need to set up an ICE table and solve for x, the percent ionization: HCO₂H ⇌ H+ + CO₂- I 0.322 M 0 0.178 M C -x +x +x E 0.322-x x 0.178+x Substituting into the expression for the ionization constant: K_a = [H+][CO₂-]/[HCO₂H] 1.77 × 10^-4 = x^2/(0.322-x) Since the percent ionization is low (less than 5%) and x is small compared to the initial concentration (0.322 M), we can make the approximation that 0.322 - x ≈ 0.322. Using this simplification, we can solve for x: 1.77 × 10^-4 = x^2/0.322 x^2 = (1.77 × 10^-4)(0.322) x = 0.1011 M Now, we can solve for the percent ionization: % ionization = (x/[HCO₂H]) x 100% % ionization = (0.1011/0.322) x 100% % ionization = 31.4% The closest answer choice is B, which is 35.6%. However, this value is too high and likely a result of assuming the initial concentration of formic acid is equal to the total concentration of formic acid and sodium formate (0.5 M), rather than just the concentration of formic acid (0.322 M). Therefore, the best choice is B with the explanation that the percent ionization is actually 31.4%, not 35.6%.

Calculate the Percent Ionization of Formic Acid (HCO2H)in a Solution

Calculate the Percent Ionization of Formic Acid (HCO₂H)in a Solution