Question

100 ml of 0.1 M H2So4 is mixed with 100 ml of 0.2 Al2(So4)3 then what will be the molar concentration of H+ ions So4 2- ions and aluminium ions in the resulting solution?

Answer 1

100ml of 0.1M H2SO4 is mixed with 100ml of 0.2M Al2(SO4)3.

number of mole of H2SO4 = concentration of H2SO4 × volume of H2SO4 in litre

= 0.1 × 100/1000 = 0.01mol

so, mole of H+ ions in H2SO4 = 2 × 0.01 = 0.02 mol

and mole of SO4²- ions in H2SO4 = 0.01 mol

similarly,

number of mole of A2(SO4)3 = 0.2M × (100/1000)L

= 0.02 mol

so, mole of SO4²- ions in Al2(SO4)3 = 3 × 0.02 = 0.06 mol

now, total number of mole of H+ ions in mixture = 0.02mol

so, concentration of H+ ions in mixture = mole of H+ ions/volume of mixture

= 0.02 × 1000/(100 + 100)

= 0.1M

and total number of mole of SO4²- ions in the mixture = 0.01 + 0.06 = 0.07mol

so, concentration of SO4²- ions in mixture = 0.07 × 1000/(100 + 100)

= 0.35M

Answer 2

Answer:

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Explanation:

100ml of 0.1M H2SO4 is mixed with 100ml of 0.2M Al2(SO4)3.

number of mole of H2SO4 = concentration of H2SO4 × volume of H2SO4 in litre

= 0.1 × 100/1000 = 0.01mol

so, mole of H+ ions in H2SO4 = 2 × 0.01 = 0.02 moland mole of SO4²- ions in H2SO4 = 0.01 mol

similarly,

number of mole of A2(SO4)3 = 0.2M × (100/1000)L

= 0.02 mol

so, mole of SO4²- ions in Al2(SO4)3 = 3 × 0.02 = 0.06 mol

now, total number of mole of H+ ions in mixture = 0.02mol

so, concentration of H+ ions in mixture = mole of H+ ions/volume of mixture

= 0.02 × 1000/(100 + 100)

= 0.1M

and total number of mole of SO4²- ions in the mixture = 0.01 + 0.06 = 0.07mol

so, concentration of SO4²- ions in mixture = 0.07 × 1000/(100 + 100)

= 0.35M