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