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Standard XII

Chemistry

Catalysis

1.575 grams o...

Question

1.575 grams of C₂H₂O₄.xH₂O is dissolved in water and its volume is made up to 250ml. On titration, 16.68ml of this solution requires 25ml of (1/15) Normal NaOH solution for complete neutralisation.
Find x.
Explain the processes and calculations.

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Solution

$Hi, The neutralisation reaction between oxalic acid and NaOH may be represented as {(COOH)}_{2} . {xH}_{2} O + NaOH = 2 COONa + (x + 2) H_{2} O 1 mole of oxalic acid requires 2 moles of NaOH 25 ml of \frac{N}{15} NaOH solution requires = 0.025 \times \frac{1}{15} mol L^{- 1} = {1.67 ×10}^{- 3} moles {1.67 ×10}^{- 3} moles of NaOH neutralise \frac{{1.67 ×10}^{- 3} moles}{2} of oxalic acid Molarity of 250 ml of oxalic acid = \frac{0.835 \times 10^{- 3} moles}{0.01668 l} = 0.05 M 0.250 L \times 0.05 M = 0.0125 moles of oxalic acid is equivalent to 1.575 g of oxalic acid . 1 mole of oxalic acid is = \frac{1.575 g}{0.0125 moles} = 126 g / mol so, Molar mass of oxalic acid {(COOH)}_{2} . {xH}_{2} O = 126 g / mol 90 + 80 x = 126 18 x = 126 - 90 = 36 x = 2 {(COOH)}_{2} . 2 H_{2} O Regards$

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1.575 grams of C2H2O4.xH2O is dissolved in water and its volume is made up to 250ml. On titration, 16.68ml of this solution requires 25ml of (1/15) Normal NaOH solution for complete neutralisation. Find x. Explain the processes and calculations.

1.575 grams of C₂H₂O₄.xH₂O is dissolved in water and its volume is made up to 250ml. On titration, 16.68ml of this solution requires 25ml of (1/15) Normal NaOH solution for complete neutralisation.
Find x.
Explain the processes and calculations.