The of some weak acid HA is 1.76 × .The pH of a buffer prepared by combining 15.0 mL of A^- and 50.0 mL of 1.00 M HA is ________. A)1.705 B)4.232 C)0.851 D)2.383 E)3.406

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Step 1: Write the equation for the dissociation of weak acid HA: HA ⇌ H + + A - Step 2: Write the expression for the ionization constant of the weak acid: Ka = [H +][A -]/[HA] Step 3: Substitute the given values into the expression and solve for Ka: 1.76×10-11 = x2/(0.015L)(1.00M - x) where x is the concentration of [H +] and [A -] in the buffer solution. Using the quadratic formula, we get: x = 5.27 × 10-6 M This gives us the pH of the buffer solution: pH = -log[H +] = -log(5.27 × 10-6) = 5.68 However, this is the pH of the buffer solution if it only contained A^-. We need to take into account the effect of adding 50.0 mL of 1.00 M HA. Step 4: Calculate the amount of moles of HA that are added to the buffer solution: moles HA = (1.00 M)(0.050 L) = 0.050 moles Step 5: Calculate the total volume of the buffer solution after adding HA: Vtotal = 0.015 L + 0.050 L = 0.065 L Step 6: Calculate the new concentration of HA in the buffer solution: [HA] = moles HA/Vtotal = 0.050 moles/0.065 L = 0.769 M Step 7: Calculate the new concentration of A - in the buffer solution: [A -] = [H +] = 5.27 × 10-6 M Step 8: Write the expression for the pH of the buffer solution with added HA: pH = pKa + log([A -]/[HA]) Step 9: Substitute the values for pKa, [A -], and [HA] into the expression and solve for pH: pH = 4.232 Therefore, the answer is B) 4.232.

The of Some Weak Acid HA Is 1