QuestionJuly 10, 2025

1. (XC5pt) Calculate the concentration of each species of X in a 0.551 M solution of Na_(2)X_(41)= 5.59e-2,K_(22)=1.96e-7 [H_(2)X]=underline ( ) [HX^-]=underline ( )[X^2-]=underline ( ) [H_(3)O^+]=underline ( ) [OH-]=underline ( )pH=underline ( )

1. (XC5pt) Calculate the concentration of each species of X in a 0.551 M solution of Na_(2)X_(41)= 5.59e-2,K_(22)=1.96e-7 [H_(2)X]=underline ( ) [HX^-]=underline ( )[X^2-]=underline ( ) [H_(3)O^+]=underline ( ) [OH-]=underline ( )pH=underline ( )
1. (XC5pt) Calculate the concentration of each species of X in a 0.551 M solution of Na_(2)X_(41)= 5.59e-2,K_(22)=1.96e-7 [H_(2)X]=underline ( ) [HX^-]=underline ( )[X^2-]=underline ( ) [H_(3)O^+]=underline ( ) [OH-]=underline ( )pH=underline ( )

Solution
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Answer

[H_2X] = 0.551 \, \text{M}, [HX^-] = 3.08 \times 10^{-3} \, \text{M}, [X^{2-}] = 1.96 \times 10^{-7} \, \text{M}, [H_3O^+] = 3.08 \times 10^{-3} \, \text{M}, [OH^-] = 3.25 \times 10^{-12} \, \text{M}, pH = 2.51 Explanation 1. Identify the dissociation reactions Na_2X_{41} dissociates into 2Na^+ and X^{2-}. The species H_2X, HX^-, and X^{2-} are involved in equilibrium reactions. 2. Write equilibrium expressions For H_2X \rightleftharpoons HX^- + H^+, use K_{a1} = 5.59 \times 10^{-2}. For HX^- \rightleftharpoons X^{2-} + H^+, use K_{a2} = 1.96 \times 10^{-7}. 3. Calculate initial concentrations Initial concentration of X^{2-} is 0.551 M from Na_2X_{41}. 4. Apply equilibrium expressions Use **K_a = \frac products reactants ** for each step to find concentrations: - [H^+] = \sqrt{K_{a1} \cdot [H_2X - [HX^-] = K_{a1} \cdot [H_2X] / [H^+] - [X^{2-}] = K_{a2} \cdot [HX^-] / [H^+] 5. Calculate [H_3O^+] and pH [H_3O^+] = [H^+]; **pH = -\log[H_3O^+]** 6. Calculate [OH^-] Use **[OH^-] = \frac{K_w} H_3O^+ **, where K_w = 1.0 \times 10^{-14}.

Explanation

1. Identify the dissociation reactions<br /> $Na_2X_{41}$ dissociates into $2Na^+$ and $X^{2-}$. The species $H_2X$, $HX^-$, and $X^{2-}$ are involved in equilibrium reactions.<br /><br />2. Write equilibrium expressions<br /> For $H_2X \rightleftharpoons HX^- + H^+$, use $K_{a1} = 5.59 \times 10^{-2}$.<br /> For $HX^- \rightleftharpoons X^{2-} + H^+$, use $K_{a2} = 1.96 \times 10^{-7}$.<br /><br />3. Calculate initial concentrations<br /> Initial concentration of $X^{2-}$ is 0.551 M from $Na_2X_{41}$.<br /><br />4. Apply equilibrium expressions<br /> Use **$K_a = \frac{[products]}{[reactants]}$** for each step to find concentrations:<br />- $[H^+] = \sqrt{K_{a1} \cdot [H_2X]}$<br />- $[HX^-] = K_{a1} \cdot [H_2X] / [H^+]$<br />- $[X^{2-}] = K_{a2} \cdot [HX^-] / [H^+]$<br /><br />5. Calculate $[H_3O^+]$ and $pH$<br /> $[H_3O^+] = [H^+]$; **$pH = -\log[H_3O^+]$**<br /><br />6. Calculate $[OH^-]$<br /> Use **$[OH^-] = \frac{K_w}{[H_3O^+]}$**, where $K_w = 1.0 \times 10^{-14}$.
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