Answer:
0.0991 M
Explanation:
Step 1: Write the neutralization reaction between oxalic acid and sodium hydroxide.
H₂C₂O₄ + 2 NaOH = Na₂C₂O₄ + 2 H₂O
Step 2: Calculate the moles of oxalic acid
The molar mass of H₂C₂O₄ is 90.03 g/mol. The moles corresponding to 153 mg (0.153 g) are:
[tex]0.153g \times \frac{1mol}{90.03g} = 1.70 \times 10^{-3} mol[/tex]
Step 3: Calculate the moles of sodium hydroxide
The molar ratio of H₂C₂O₄ to NaOH is 1:2.
[tex]1.70 \times 10^{-3} molH_2C_2O_4 \times \frac{2molNaOH}{1molH_2C_2O_4} = 3.40 \times 10^{-3} molNaOH[/tex]
Step 4: Calculate the molarity of sodium hydroxide
[tex]\frac{3.40 \times 10^{-3} mol}{34.3 \times 10^{-3} L} = 0.0991 M[/tex]
