# Question 4.5: GRADED Three beakers labeled A, B, and C contain the weak ac...

Three beakers labeled A, B, and C contain the weak acid $H_{2}X$. The weak acid is titrated with 0.125 M NaOH. Assume the reaction to be

$H_{2}X$(aq) + 2 $OH^{-}$(aq) → 2$H_{2}O$ + $X^{2-}$(aq)

ⓐ Beaker A contains 25.00 mL of 0.316 M $H_{2}X$. What volume of NaOH is required for complete neutralization?

ⓑ Beaker B contains 25.00 mL of a solution of $H_{2}X$ and requires 28.74 mL of NaOH for complete neutralization.

ⓒ Beaker C contains 0.124 g of $H_{2}X$ and 25.00 mL of water. To reach the equivalence point, 22.04 mL of NaOH are required. What is the molar mass of $H_{2}X$?

 ANALAYSIS volume (25.00 mL) and molarity (0.316 M) of $H_{2}X$ molarity (0.125 M) of NaOH net ionic equation [$H_{2}X$(aq) + 2$OH^{-}$(aq) → 2$H_{2}O$ + $X^{2-}$(aq)] Information given: stoichiometric ratio; reacting species Information implied: volume of NaOH required for neutralization Asked for:

STRATEGY

1. Use the stoichiometric ratio: 2 mol $OH^{-}$/1 mol $H_{2}X$
2. Follow the flow chart in Figure 4.6.
$H_{2}X$ does not break up into ions. Skip the moles parent compound → moles ion step

mol $H_{2}X \overset{stoichiometric }{\underset{ratio }{\longrightarrow }} mol OH^{-} \longrightarrow mol NaOH \underrightarrow{n \div M} volume NaOH$

 ANALAYSIS volume (28.74 mL) and molarity (0.125 M) of NaOH volume of $H_{2}X$ (25.00 mL) required for complete neutralization net ionic equation [$H_{2}X$(aq) + 2$OH^{-}$(aq) → 2$H_{2}O$ + $X^{2-}$(aq)] Information given: stoichiometric ratio Information implied: molarity of $H_{2}X$ Asked for:

STRATEGY

1. Use the stoichiometric ratio: 2 mol $OH^{-}$/1 mol $H_{2}X$
2. Follow the flow chart in Figure 4.6.
$H_{2}X$ does not break up into ions. Skip the moles parent compound → moles ion step for $H_{2}X$.

mol NaOH $\longrightarrow mol OH^{-} \overset{stoichiometric }{\underset{ratio }{\longrightarrow }} mol H_{2}X \underrightarrow{n \div V} M_{H_{2}X}$

 ANALAYSIS volume (22.04 mL) and molarity (0.125 M) of NaOH mass (0.124 g) of H2X volume (25.00 mL) of water net ionic equation [$H_{2}X$(aq) + 2$OH^{-}$(aq) → 2$H_{2}O$ + $X^{2-}$(aq)] Information given: stoichiometric ratio Information implied: molar mass of $H_{2}X$ Asked for:

STRATEGY

1. Use the stoichiometric ratio: 2 mol $OH^{-}$/1 mol $H_{2}X$
2. Follow the flow chart in Figure 4.6.
$H_{2}X$ does not break up into ions. Skip the moles parent compound : moles ion step for $H_{2}X$.

mol NaOH $\longrightarrow mol OH^{-} \overset{stoichiometric }{\underset{ratio }{\longrightarrow }} mol H_{2}X \underrightarrow{mass \div n} MM of H_{2}X$

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