A buffer is 0.282 M C₆H₅COOH(aq)and 0.282 M C₆H₅COONa(aq).Calculate the pH after the addition of 0.150 moles of nitric acid to 1.0 L of the buffer.For C₆H₅COOH,pK_a = 4.20. A)3.69 B)4.20 C)4.71 D)3.87 E)7.00

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The correct answer is A because the addition of nitric acid will decrease the pH of the buffer. The pH of the buffer can be calculated using the Henderson-Hasselbalch equation:pH = pKa + log([A-]/[HA])where pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.In this case, the pKa of C6H5COOH is 4.20, and the initial concentrations of C6H5COOH and C6H5COONa are both 0.282 M. Therefore, the initial pH of the buffer is:pH = 4.20 + log([0.282]/[0.282]) = 4.20After the addition of 0.150 moles of nitric acid to 1.0 L of the buffer, the concentration of C6H5COOH will increase and the concentration of C6H5COONa will decrease. The new concentrations can be calculated using the following equations:[C6H5COOH] = 0.282 M + 0.150 moles / 1.0 L = 0.432 M[C6H5COONa] = 0.282 M - 0.150 moles / 1.0 L = 0.132 MTherefore, the new pH of the buffer is:pH = 4.20 + log([0.132]/[0.432]) = 3.69

A Buffer Is 0