QuestionJune 1, 2025

(9) Given a pure 0.50M HCN solution calculate the volume that when diluted to a total volume of 250.0 mL will yield a pH that is 0.65 unit higher than that of the given solution. Consider the volumes of the original solution and of the diluent to be precisely additive.

(9) Given a pure 0.50M HCN solution calculate the volume that when diluted to a total volume of 250.0 mL will yield a pH that is 0.65 unit higher than that of the given solution. Consider the volumes of the original solution and of the diluent to be precisely additive.
(9) Given a pure 0.50M HCN solution calculate the volume that when diluted to a total volume of 250.0 mL will yield a pH that is 0.65 unit higher than that of the given solution. Consider the volumes of the original solution and of the diluent to be precisely additive.

Solution
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Answer

The volume of the original 0.50M HCN solution needed is approximately 7.8 mL. Explanation 1. Calculate initial pH of 0.50M HCN Use the formula for weak acid dissociation: K_a = \frac H^+][CN^- HCN . For HCN, K_a = 6.2 \times 10^{-10}. Assume [H^+] = x, then x^2 = K_a \cdot [HCN]. Solve for x: x = \sqrt{6.2 \times 10^{-10} \times 0.50}. Calculate pH = -\log_{10}(x). 2. Determine new pH after dilution New pH is 0.65 units higher than initial pH. If initial pH is pH_1, then new pH = pH_1 + 0.65. 3. Calculate new [H^+] concentration Use [H^+] = 10^{-\text{new pH}} to find new [H^+] concentration. 4. Calculate new concentration of HCN Use K_a = \frac H^+]^2} HCN to solve for new [HCN]: [HCN] = \frac H^+]^2}{K_a}. 5. Calculate volume of original solution needed Use dilution formula: C_1V_1 = C_2V_2. Here, C_1 = 0.50M, V_2 = 250.0 \, \text{mL}, and C_2 is the new concentration from Step 4. Solve for V_1: V_1 = \frac{C_2 \times V_2}{C_1}.

Explanation

1. Calculate initial pH of 0.50M HCN<br /> Use the formula for weak acid dissociation: $K_a = \frac{[H^+][CN^-]}{[HCN]}$. For HCN, $K_a = 6.2 \times 10^{-10}$. Assume $[H^+] = x$, then $x^2 = K_a \cdot [HCN]$. Solve for $x$: $x = \sqrt{6.2 \times 10^{-10} \times 0.50}$. Calculate $pH = -\log_{10}(x)$.<br /><br />2. Determine new pH after dilution<br /> New pH is 0.65 units higher than initial pH. If initial pH is $pH_1$, then new pH $= pH_1 + 0.65$.<br /><br />3. Calculate new $[H^+]$ concentration<br /> Use $[H^+] = 10^{-\text{new pH}}$ to find new $[H^+]$ concentration.<br /><br />4. Calculate new concentration of HCN<br /> Use $K_a = \frac{[H^+]^2}{[HCN]}$ to solve for new $[HCN]$: $[HCN] = \frac{[H^+]^2}{K_a}$.<br /><br />5. Calculate volume of original solution needed<br /> Use dilution formula: $C_1V_1 = C_2V_2$. Here, $C_1 = 0.50M$, $V_2 = 250.0 \, \text{mL}$, and $C_2$ is the new concentration from Step 4. Solve for $V_1$: $V_1 = \frac{C_2 \times V_2}{C_1}$.
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