Holooly Plus Logo

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

fig 4.6
The "Step-by-Step Explanation" refers to a detailed and sequential breakdown of the solution or reasoning behind the answer. This comprehensive explanation walks through each step of the answer, offering you clarity and understanding.
Our explanations are based on the best information we have, but they may not always be right or fit every situation.
The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.
Already have an account?

Related Answered Questions

Question: 4.8

Verified Answer:

N: +5 → +2; O: -2 → -2; N is reduced. (1) (a) Oxid...