A buffer is prepared by adding 1.50 × 10² mL of 0.250 M NaOH to 2.50 × 10² mL of 0.350 M weak acid,HB.The solution is diluted to 1.00 L.If the pH of the resulting solution is 6.55,what is the pK_a of the weak acid?

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The correct answer is D.The pH of the buffer solution can be calculated using the Henderson-Hasselbalch equation:pH = pKa + log([A-]/[HA])where [A-] is the concentration of the conjugate base and [HA] is the concentration of the weak acid.We can use the initial concentrations of NaOH and HB to calculate the initial concentrations of [A-] and [HA]:[A-] = (1.50 × 10² mL)(0.250 M) / 1.00 L = 0.0375 M[HA] = (2.50 × 10² mL)(0.350 M) / 1.00 L = 0.0875 MSubstituting these values into the Henderson-Hasselbalch equation, we get:6.55 = pKa + log(0.0375/0.0875)pKa = 6.92

A Buffer Is Prepared by Adding 1