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## Q. 4.5

GRADED

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

## Verified Solution

 0.02500 L × 0.316 $\frac{mol H_{2}X}{L}$ × $\frac{2 mol OH^{-}}{1 mol H_{2}X}$ × $\frac{1 mol NaOH}{1 mol OH^{-}}$ = 0.0158 mol NaOH V = n ÷ M = $\frac{0.0158 mol}{0.125 M}$ = 0.126 L Volume of NaOH used

 0.02874 L × 0.125 $\frac{mol NaOH}{L}$ × $\frac{1 mol OH^{-}}{1 mol NaOH}$ × $\frac{1 mol H_{2}X}{2 mol OH^{-}}$ = 0.00180 Mol $H_{2}X$ M = n ÷ M = $\frac{0.00180 mol}{0.02500 L}$ = 0.0720  M Molarity of $H_{2}X$ (M)

 0.02204 L × 0.125 $\frac{mol NaOH}{L}$ × $\frac{1 mol OH^{-}}{1 mol NaOH}$ × $\frac{1 mol H_{2}X}{2 mol OH^{-}}$ = 0.001378 mol $H_{2}X$ MM = mass ÷ n = $\frac{0.124 g}{0.001378 mol}$ = 90.0  g/mol molar mass of $H_{2}X$

END POINT

1. You need to figure out the number of moles before you can calculate mass, molar mass, volume, or molarity.
2. The amount of water added to the solid $H_{2}X$ is irrelevant to the solution of the problem.