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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

fig 4.6

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.