Multiple Choice
Which of the following solutions will have the lowest concentration of hydronium ions?
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17. Acid and Base Equilibrium
The pH Scale
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Determine the pH of a solution made by dissolving 6.1 g of sodium cyanide, NaCN, in enough water to make a 500.0 mL of solution. (MW of NaCN = 49.01 g/mol). The Ka value of HCN is 4.9 x 10-10.
A
9.04
B
4.96
C
11.35
D
2.65
Verified step by step guidance1
Calculate the number of moles of NaCN by dividing the mass of NaCN (6.1 g) by its molar mass (49.01 g/mol). Use the formula: \( \text{moles of NaCN} = \frac{6.1 \text{ g}}{49.01 \text{ g/mol}} \).
Determine the concentration of NaCN in the solution by dividing the number of moles of NaCN by the volume of the solution in liters (0.500 L). Use the formula: \( \text{[NaCN]} = \frac{\text{moles of NaCN}}{0.500 \text{ L}} \).
Recognize that NaCN is a salt that dissociates completely in water to form Na\(^+\) and CN\(^-\). The CN\(^-\) ion is the conjugate base of the weak acid HCN, and it will undergo hydrolysis in water to form OH\(^-\) ions.
Use the hydrolysis equation for CN\(^-\): \( \text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^- \). Set up an expression for the base dissociation constant \( K_b \) using the relation \( K_w = K_a \times K_b \), where \( K_w = 1.0 \times 10^{-14} \) and \( K_a = 4.9 \times 10^{-10} \). Solve for \( K_b \).
Calculate the concentration of OH\(^-\) ions using the expression for \( K_b \) and the initial concentration of CN\(^-\). Use the formula: \( K_b = \frac{[\text{HCN}][\text{OH}^-]}{[\text{CN}^-]} \). Then, determine the pOH of the solution and convert it to pH using the relation \( \text{pH} + \text{pOH} = 14 \).
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