The number of Hydrogen ions in 1.0{dm}³ of 0.02{mol\,dm}^{-3} tetraoxosulphate(VI) acid is ≤ft[ N_A = 6.02 × 10²3} ]

Correct Answer is : 2.4×10222.4 × 10²2}2.4×1022

Correct Answer is : 2.4×10222.4 × 10²2}2.4×1022

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WAEC Chemistry 2018 Paper

Question : 21

Total: 50

\mathrm{dm}^{3}

\mathrm{mol\,dm}^{-3}

\left[ N_A = 6.02 \times 10^{23} \right]

Solution:

Concept:

Tetraoxosulphate(VI) acid is

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

, a dibasic acid that releases

2\,\mathrm{H}^+

ions per molecule.

Explanation:

Moles of

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

in the solution is:

n = 0.02\,\mathrm{mol\,dm}^{-3} \times 1.0\,\mathrm{dm}^3 = 0.02\,\mathrm{mol}

Since each mole of

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

produces

2\,\mathrm{mol}

\mathrm{H}^+

ions:

\text{moles of }\mathrm{H}^+ = 2 \times 0.02 = 0.04\,\mathrm{mol}

Number of

\mathrm{H}^+

ions is:

0.04 \times 6.02 \times 10^{23} = 2.408 \times 10^{22}

Rounded to two significant figures:

= 2.4 \times 10^{22}

Answer:

Option C:

2.4 \times 10^{22}

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