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

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The correct answer is A because the pH of a buffer can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa is the negative logarithm of 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 HCN is 6.2 x 10^-10, [A-] is 0.53 M, and [HA] is 0.88 M. Plugging these values into the equation, we get pH = 6.2 x 10^-10 + log(0.53/0.88) = 8.99.

What Is the PH of a Buffer That Is 0