InterviewSolution
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InterviewSolution
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10 " mL of " 2 M HCl and 20 " mL of " 1 M `HNO_3` and V volume of `5M H_2SO_4` are mixed together and the solution was made upto 5 L. 10 " mL of " this acid solution exactly neutralises 28.6 " mL of " `Na_2CO_3` solution containing 1 g of `Na_2CO_4.10H_2O` in 100 " mL of " water. Calculate the amount of `SO_4^(2-)` ions in grams present in solution. |
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Answer» Mw of `Na_2CO_3.10H_2O=2xx23+12+3xx16+10xx18=286g` Ew of `Na_2CO_3.10H_2O("valency factor"=2)=(286)/(2)=143g` Strength of `Na_2CO_3.10H_2O=(1xx1000)/(100)=10gL^(-1)` `N_(Na_(2)CO_(3))=("Strength")/(Ew)=(10)/(143)` total mEq of mixture of acid in 5 L of solution`=10xx2xx1+20xx1xx1xxVxx5xx2` (n-factor) `=20+20+10V=40+10V` Now m" Eq of "acid in 10 mL solution `=m" Eq of "Na_2CO_3` used for it `=N_(Na_(2)CO_(3))xxV(mL)=(10)/(143)xx28.6` `=2mEq(i n 10mL)` `=(2xx5000)/(10)mEq(i n 5L)` `=1000mEq` m" Eq of "acid in 5 L solution`=1000mEq` `40+10V=1000` `10V=960` `10V-=960-=m" Eq of "H_2SO_4-=m" Eq of "SO_4^^(2-)` `m" Eq of "SO_4^(2-)=(weight)/(Ew)xx10^(3)=960` `(weightxx10^(3))/((96)/(2))=960` Weight of `SO_4^(2-)=(960xx48)/(1000)=4.6g` |
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