Chapter 4
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) → 2H_{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) + 2OH^{-}(aq) → 2H_{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) + 2OH^{-}(aq) → 2H_{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) + 2OH^{-}(aq) → 2H_{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
Step-by-Step
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.