1 mol each of ${H}_3 {PO}_2, {H}_3 {PO}_3$ and ${H}_3 {PO}_4$ will neutralize $x \mathrm{~mol}$ of ${NaOH}, y \mathrm{~mol}$ of ${Ca}({OH})_2$ and ${z} \mathrm{~mol}$ of ${Al}({OH})_3 . x, y$ and $z$ are in the ratio

JEE CHEMISTRY MCQ: SOME BASIC CONCEPTS OF CHEMISTRY

1 Hard

1 of 1

Application +4 | -1

1 mol each of

{H}_3 {PO}_2, {H}_3 {PO}_3

and

{H}_3 {PO}_4

will neutralize

x \mathrm{~mol}

{NaOH}, y \mathrm{~mol}

{Ca}({OH})_2

and

{z} \mathrm{~mol}

{Al}({OH})_3 . x, y

and

z

are in the ratio

View answer and explanation

Correct option: D

1: 1: 1

Option B (

4.78 \times 10^{-11} \mathrm{~mol} \mathrm{~L}^{-1}

) is correct because it is the concentration of CO3-2 calculated using the given pH, K1, and K2 values for carbonic acid.

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{H_2}O.

mol\,\,k{g^{ - 1}})

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{CO}_2

0.5 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4

0.5 \mathrm{~g}

\mathrm{Cu}(\mathrm{II})