The reaction of oxalic acid and sodium hydroxide solution produces oxalate ion, C2O4^-2, as one of the products.
A) Write the balanced net ionic equation for the reaction of sodium hydroxide solution with oxalic acid solution.
B) Calculate the mass(g) of oxalic acid dihydrate required to prepare 250.0mL of 0.0300N H2C2O4.
The Correct Answer and Explanation is :
A) Balanced Net Ionic Equation
The reaction between oxalic acid ((H_2C_2O_4)) and sodium hydroxide ((NaOH)) is a neutralization reaction. Oxalic acid is diprotic, meaning it can donate two protons ((H^+)) to form the oxalate ion ((C_2O_4^{2-})):
- The first step involves the loss of the first proton:
[
H_2C_2O_4 \leftrightarrow H^+ + HC_2O_4^-
] - The second step involves the loss of the second proton:
[
HC_2O_4^- \leftrightarrow H^+ + C_2O_4^{2-}
]
The reaction with (NaOH) neutralizes these protons. The net ionic equation is:
[
H_2C_2O_4(aq) + 2OH^-(aq) \rightarrow C_2O_4^{2-}(aq) + 2H_2O(l)
]
B) Mass of Oxalic Acid Dihydrate
To calculate the mass of oxalic acid dihydrate ((H_2C_2O_4 \cdot 2H_2O)) required to prepare 250.0 mL of 0.0300 N solution:
- Determine the equivalent weight of oxalic acid dihydrate:
- Molar mass of (H_2C_2O_4): (2(1.01) + 2(12.01) + 4(16.00) = 90.03 \, \text{g/mol})
- Molar mass of (2H_2O): (2(18.02) = 36.04 \, \text{g/mol})
- Total molar mass: (90.03 + 36.04 = 126.07 \, \text{g/mol})
- Since oxalic acid is diprotic, ( \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{Number of acidic protons}} = \frac{126.07}{2} = 63.04 \, \text{g/equiv}).
- Calculate the mass required:
- Normality ((N)) relates to equivalents per liter:
[
N = \frac{\text{equivalents}}{\text{volume (L)}}
] - Equivalents of oxalic acid required:
[
0.0300 \, \text{N} \times 0.2500 \, \text{L} = 0.00750 \, \text{equiv}
] - Mass required:
[
\text{Mass} = \text{Equivalents} \times \text{Equivalent weight}
]
[
\text{Mass} = 0.00750 \, \text{equiv} \times 63.04 \, \text{g/equiv} = 0.4728 \, \text{g}
]
Explanation (300 words)
The neutralization of oxalic acid by sodium hydroxide is a two-step reaction due to the diprotic nature of oxalic acid. Each molecule of (H_2C_2O_4) can donate two protons ((H^+)), reacting with two hydroxide ions ((OH^-)) to form the oxalate ion ((C_2O_4^{2-})) and water. The balanced net ionic equation reflects this stoichiometry.
For part B, the preparation of 0.0300N oxalic acid solution involves knowing its equivalent weight, which accounts for the two acidic protons. The molecular weight of oxalic acid dihydrate ((H_2C_2O_4 \cdot 2H_2O)) includes the dihydrate’s water content, which must be factored into the calculation.
Normality is used instead of molarity because oxalic acid is diprotic, and normality accounts for the number of protons involved in the reaction. Multiplying the normality by the solution volume gives the total equivalents of (H_2C_2O_4) needed. Converting equivalents to mass using the equivalent weight determines the precise quantity of oxalic acid dihydrate.
This method ensures accuracy in solution preparation, critical in titrations and other chemical analyses where precise concentrations are necessary.