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 × 10^-4) A) −0.301 B) 3.05 C) 3.44 D) 3.74 E) 4.05

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Step 1: Write the balanced equation for the dissociation of NaHCO₂: NaHCO₂(aq) + H2O(l) <--> HCO3- (aq) + Na+ (aq) + H3O+ (aq) Step 2: Calculate the initial concentration of NaHCO₂: 0.40 mol / (250 mL) = 0.0016 M Step 3: Determine which species will react with HCl - in this case, it is the HCO3- ion: HCO3- (aq) + HCl (aq) --> H2CO3 (aq) + Cl- (aq) Step 4: Calculate the amount of HCl that will react: 100.0 mL * 2.00 mol/L = 0.200 mol Step 5: Calculate the amount of HCO3- ion that will react, assuming all of the HCl is used up: 0.200 mol * (1 mol HCO3-/1 mol HCl) = 0.200 mol HCO3- Step 6: Calculate the final concentration of HCO3- ion: 0.0016 M - 0.200 mol / (250 mL) = 0.0008 M Step 7: Calculate the final concentration of H3O+ ion (from the dissociation of H2CO3): K_a = [HCO3-][H3O+] / [H2CO3] 1.8 × 10^-4 = (0.0008 M)([H3O+]) / [H2CO3] Assuming that the initial concentration of H2CO3 can be approximated by the final concentration of HCO3-: [H2CO3] = 0.0008 M [H3O+] = K_a * [H2CO3] / [HCO3-] = (1.8 × 10^-4)(0.0008 M) / (0.0008 M) = 1.8 × 10^-4 Step 8: Calculate the pH: pH = -log[H3O+] = -log(1.8 × 10^-4) = 3.74 Therefore, the answer is D (3.74).

What Is the PH of a Buffer That Results When