What is the pH of a buffer that is 0.88 M HCN(aq)and 0.53 M NaCN(aq)? The K_a of HCN is 6.2 × 10^-10. A)8.99 B)9.21 C)9.43 D)4.79

Question

Quizplus · Accepted Answer

The correct answer is A.The pH of a 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 acid is HCN and the conjugate base is CN-. The pKa of HCN is 6.2 × 10^-10. The concentration of HCN is 0.88 M and the concentration of CN- is 0.53 M.Substituting these values into the Henderson-Hasselbalch equation, we get:pH = 6.2 × 10^-10 + log(0.53/0.88)= 6.2 × 10^-10 + log(0.6)= 6.2 × 10^-10 - 0.22= 5.98Therefore, the pH of the buffer is 5.98.

What Is the PH of a Buffer That Is 0