Chemistry - Redox Reactions Question with Solution | TestHub

${\mathrm{P}}_4{\mathrm{O}}_{10}$ + $6{\mathrm{H}}_2\mathrm{O}$ $\to$ $4{\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4$. Moles ${\mathrm{P}}_4{\mathrm{O}}_{10}$ = 2.84/284 = 0.01, so ${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4$ = 0.04 mol in 500 ml = 0.08 M: (a) true. (b) 25 ml has 2 mmol ${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4$; complete neutralization needs 6 mmol $\mathrm{NaOH}$ = 60 ml $x$ 0.1 M - true. (c) 10 ml has 0.8 mmol phosphate; $2{\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4$ + $3BaC{\mathrm{l}}_2$ $\to$ $\mathrm{B}{\mathrm{a}}_3(\mathrm{P}{\mathrm{O}}_4{)}_2$ needs 1.2 mmol $\mathrm{BaC}{\mathrm{l}}_2$ = 12 ml $x$ 0.1 M - true. (d) 50 ml has 4 mmol; first equivalent point (to $\mathrm{K}{\mathrm{H}}_2\mathrm{P}{\mathrm{O}}_4$) needs 4 meq $\mathrm{KOH}$ $\to$ 50 ml of 0.08 N, not 40 ml - false. Hence (a), (b), (c).