A solution contains H3PO4 and NaH2PO4. Titration with 0.100 M NaOH needs 12.0 mL to reach the methyl orange end point (first stage, H3PO4 -> H2PO4-) and a total of 31.4 mL to reach the phenolphthalein end point (to HPO4^2-). Calculate the mass of NaH2PO4 in the sample, in mg, taking its molar mass as 120 g mol^-1 (nearest integer).

JEE Advanced levelNumerical answerDifficulty 3 of 3eq051

A solution contains $H X_{3} PO X_{4}$ and $NaH X_{2} PO X_{4}$ . Titration with $0.100 M$ $NaOH$ needs $12.0 mL$ to reach the methyl orange end point (first stage, $H X_{3} PO X_{4} H X_{2} PO X_{4} X^{-}$ ) and a total of $31.4 mL$ to reach the phenolphthalein end point (to $HPO X_{4} X^{2 -}$ ). Calculate the mass of $NaH X_{2} PO X_{4}$ in the sample, in mg, taking its molar mass as $120 g mo l^{- 1}$ (nearest integer).

Numerical answer type, answer in mg, tolerance 0.5.

Methyl orange step:

H X_{3} PO X_{4} = 0.100 \times 12.0 = 1.20

mmol (only its first

H X^{+}

is neutralised). To phenolphthalein, total

NaOH = 0.100 \times 31.4 = 3.14

mmol

= 2 [H X_{3} PO X_{4}] + [NaH X_{2} PO X_{4}]

. So

[NaH X_{2} PO X_{4}] = 3.14 - 2 \times 1.20 = 0.74

mmol, mass

= 0.74 \times 120 = 88.8

mg.