Galactose reacts with periodate, producing $H C O O H, H C H O$ and $I O_{3}^{-}$ only
In a typical experiment, a10mL solution of Galactose required 25mL0.75M periodate solution to reach the equivalence point. The solution is made free from formic acid and iodate ion by extraction and then treated with $H_{2} O_{2}$ , which oxidizes all formaldehyde into formic acid which is titrated against 0.1M NaOH solution. Titration required 37.5mL of alkali to reach the equivalence point. Then the molar ratio of formic acid to formaldehyde produced in the original reaction is______.

Question

Galactose reacts with periodate, producing $H C O O H, H C H O$ and $I O_{3}^{-}$ only
In a typical experiment, a10mL solution of Galactose required 25mL0.75M periodate solution to reach the equivalence point. The solution is made free from formic acid and iodate ion by extraction and then treated with $H_{2} O_{2}$ , which oxidizes all formaldehyde into formic acid which is titrated against 0.1M NaOH solution. Titration required 37.5mL of alkali to reach the equivalence point. Then the molar ratio of formic acid to formaldehyde produced in the original reaction is______.

Accepted Answer

The given redox reaction can be balanced as $C_{6} H_{12} O_{6} (a q) + x I O_{4}^{-} (a q) \to x H C O O H + (6 - x) H C H O + I O_{3}^{-}$
$∵$ X mole periodate combines with 1.0 mole galactose .
$∴ 25 \times 0.75 = 18.75$ Mmol periodate will combine with $\frac{18.75}{x}$ mmol of galactose
Also 1.0mmol of galactose produces (6-x) mole of formaldehyde. $\frac{18.75}{x}$ Mmol of galactose
$(6 - x) \times \frac{18.75}{x}$ mmol of formaldehyde.
Thus, mmol of formic acid produced from oxidation of formaldehyde $= (6 - x) x \frac{18.75}{x} = 3.75$
$\Rightarrow x = 5$ .Therefore, mmol of galactose $= 3.75$ , and molarity $= 0.375 M$ .
$\Rightarrow$ Molar ratio of HCOOH: HCHO $= 5 : 1$

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