What is the pH of a buffer that results when 0.40 mol NaHCO₂ is mixed with 100.0 mL of 2.00 M HCl(aq)and diluted with water to 250 mL? (K_a of HCO₂H = 1.8 $\times$ 10^-4) A) (-0.301) B) 3.05 C) 3.44 D) 3.74 E) 4.05

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1. Write the equation for the dissociation of NaHCO₂ in water: NaHCO₂(aq) ⇌ Na1U1B+1U1B(aq) + HCO₂H(aq) 2. Write the equation for the reaction between HCO₂H and HCl: HCO₂H(aq) + HCl(aq) ⇌ H2O(l) + NaCl(aq) + CO2(g) 3. Calculate the moles of HCl added to the buffer solution: moles HCl = Molarity x Volume = 2.00 M x 0.100 L = 0.200 mol 4. Calculate the initial moles of NaHCO₂ in the buffer solution: moles NaHCO₂ = 0.40 mol 5. Calculate the final moles of NaHCO₂ and HCO₂H in the buffer solution after the reaction with HCl: moles HCO₂H = moles NaHCO₂ - moles HCl = 0.40 mol - 0.200 mol = 0.200 mol moles NaHCO₂ = moles Na1U1B+1U1B = 0.200 mol 6. Calculate the concentration of HCO₂H and Na1U1B+1U1B in the final buffer solution: [HCO₂H] = moles HCO₂H / volume final = 0.200 mol / 0.250 L = 0.800 M [Na1U1B+1U1B] = moles Na1U1B+1U1B / volume final = 0.200 mol / 0.250 L = 0.800 M 7. Use the dissociation constant K_a of HCO₂H to calculate the pH of the buffer: K_a = [H1U1B+1U1B][CO₂H1U1B-1U1B] / [HCO₂H] = (x)(0.800) / (0.800) Solving for x, we get: x = 1.8 x 10^-4 pH = -log[H1U1B+1U1B] = -log(1.8 x 10^-4) = 3.74 Therefore, the answer is D, 3.74.

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