Q. 64.2( 81 Votes )

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?

Answer :

Molarity of HCl = 0.1 M Na2CO3 and NaHCO3 = 1 g
The weight of mixture of 
Let comma space weight space of space Na subscript 2 CO subscript 3 space present space in space mixture space equals space straight x space straight g  therefore space space weight space of space NaHCO subscript 3 space present space in space mixture space equals space open parentheses 1 minus straight x close parentheses space straight g  Molecular space weight space of space Na subscript 2 CO subscript 3 space equals 106 space straight g divided by mol  Molecular space weight space of space NaHCO subscript 3 space equals space 84 space straight g divided by mol  No. space of space moles space of space Na subscript 2 CO subscript 3 space equals straight x over 106  No. space of space moles space of space NaHCO subscript 3 space equals fraction numerator 1 minus straight x over denominator 84 end fraction  As space mixture space of space Na subscript 2 CO subscript 3 space and space NaHCO subscript 3 space containing space equimolar space amounts space of space both comma  straight x over 106 equals fraction numerator 1 minus straight x over denominator 84 end fraction  84 straight x space equals 106 minus 106 straight x  84 straight x space plus space 106 straight x space equals space 106  190 straight x space equals space 106  straight x space equals 106 over 190  space space space equals 0.5578 space straight g  Weight space of space Na subscript 2 CO subscript 3 space present space in space mixture space equals 0.5578 space straight g  therefore space No. space of space moles space of space Na subscript 2 CO subscript 3 space equals fraction numerator 0.5578 over denominator 106 end fraction  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.00526 space moles  space Weight space of space NaHCO subscript 3 space present space in space mixture space equals space 1 minus 0.5578  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.4422 space straight g   No. space of space moles space of space NaHCO subscript 3 space equals fraction numerator 0.4422 over denominator 84 end fraction  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.00526 space moles  Now comma  Na subscript 2 CO subscript 3 space plus space 2 HCl space rightwards arrow space 2 NaCl space space plus space straight H subscript 2 straight O space plus space CO subscript 2  space NaHCO subscript 3 space plus space HCl space rightwards arrow space NaCl space plus space straight H subscript 2 straight O space plus space CO subscript 2  So comma space 1 space mole space of space Na subscript 2 CO subscript 3 space requires space 2 space moles space of space HCl  therefore 0.00526 space moles space of space Na subscript 2 CO subscript 3 space will space require space fraction numerator 0.00526 cross times 2 over denominator 1 end fraction  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.0105 space moles  1 space mole space of space space NaHCO subscript 3 space requires space 1 space moles space of space HCl  therefore space space 0.00526 space moles space of space NaHCO subscript 3 space will space require space space fraction numerator 1 cross times 0.00526 over denominator 1 end fraction  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.00526 space moles  Tota semicolon space HCl space required space equals space 0.0105 plus space 0.00526  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 0.01578 space moles space space space space space space space space space space space space space space space space space space space space space space space space space space space space   0.1 space straight M space HCl space means  space 0.1 space mole space of space HCl space in space 1000 space ml  therefore space 0.01578 space moles space of space HCl space will space present space in space space fraction numerator 0.01578 cross times 1000 over denominator 0.1 end fraction space  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 157.88 space ml
The volume HCl required is 157.88 ml.

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