What mass of sodium cyanide must be added to 250. mL of water in order to obtain a solution having a pH of 10.50? [K_a(HCN) = 4.9 * 10^-10] A) 240 g B) 0.032 g C) 0.059 g D) 0.94 g E) 0.24 g

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1. Write out the chemical equation of sodium cyanide in water: NaCN + H2O → Na+ + CN- + H2O. 2. The equilibrium constant for this reaction can be expressed as: K_a(HCN) = [CN-][H2O]/[NaCN]. 3. Since we want a solution with a pH of 10.50, we know that [H+] = 10^-10.50 = 3.16 × 10^-11 M. 4. At equilibrium, the concentration of CN- will be equal to the concentration of HCN (cyanide ion is the conjugate base of HCN). 5. We can use the pH and the pKa of HCN (9.21) to calculate the concentration of HCN at equilibrium: [HCN] = [H+] / [CN-] = 3.16 × 10^-11 M / 10^-0.29 = 5.89 × 10^-3 M. 6. Now we can use the equilibrium constant expression to solve for the concentration of NaCN: K_a(HCN) = [CN-][H2O]/[NaCN] = (5.89 × 10^-3 M)(55.6 M) / [NaCN] = 4.9 × 10^10. 7. Solve for [NaCN]: [NaCN] = (5.89 × 10^-3 M)(55.6 M) / 4.9 × 10^10 = 6.70 × 10^-9 M. 8. Calculate the mass of NaCN needed: Mass = (6.70 × 10^-9 mol/L)(0.250 L)(49.01 g/mol) = 0.059 g. Therefore, the answer is C.

What Mass of Sodium Cyanide Must Be Added to 250