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Calculate the molarityes and normalities of the solution obtained on mixing (i) 100 mL of 0.2 M H_(2)SO_(4) with 50 mL of 0.1 M HCl (ii) 100 mL of 0.2 N H_(2)SO_(4) with 50 mL of 0.1 N HCl (iii) 100 mL of 0.1 M H_(2)SO_(4) with 50 mL of 0.1 M NaOH (iv) 50 mL of 0.1 N H_(2)SO_(4) with 100 mL of 0.1 N NaOH. |
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Answer» Solution :(i) `"100 mL of 0.2 M "H_(2)SO_(4)" contain "H_(2)SO_(4)=(0.2)/(1000)xx100=0.02mol="0.02 g eq."` `"50 mL of 0.1 M HCl contain HC"=(0.1)/(1000)xx50="0.005 mol = 0.005 g eq."` `"Total no. of MOLES present "=0.02+0.005mol=0.025 mol` `"Total VOLUME after mixing "=100+50mL=150mL=0.150L` `"Molarity"=(0.025mol)/(0.150 L)=0.167M` `"Total no. of g eq."=0.04+0.005="0.045g eq."` `"Normality"=("0.045 g eq")/("0.150 L")=0.3N` Alternatively, `M_(1)V_(1)+M_(2)V_(2)=M_(3)(V_(1)+V_(2))` `therefore""M_(3)=(0.2xx100+0.1xx50)/(150)=0.167M` Similarly, `N_(1)V_(1)+N_(2)V_(2)=N_(3)(V_(1)+V_(2))` `"As0.2 M "H_(2)SO_(4)="0.4 N "H_(2)SO_(4) and " 0.1 M HCl = 0.1 N HCl"` `N_(3)=(0.4xx100+0.1xx50)/(150)=0.3N` (ii) `"0.2 N "H_(2)SO_(4)="0.1 M "H_(2)SO_(4)"and0.1 N HCl = 0.1 M HCl"` `"Applying"M_(1)V_(1)+M_(2)V_(2)=M_(3)(V_(1)+V_(2))` `M_(3)=(0.1xx100+0.1xx50)/(150)=(15)/(150)=0.1M` `"Applying"N_(1)V_(1)+N_(2)V_(2)=N_(3)(V_(1)+V_(2))` `N_(3)=(0.2xx100+0.1xx50)/(150)=(25)/(150)=0.167N` (III) `"100 mL of 0.1 M "H_(2)SO_(4)" contain "H_(2)SO_(4)=(0.1)/(1000)xx100="0.01 mol = 0.02 g eq"` `"50 mL of 0.1 M NAOH contain NaOH"=(0.1)/(1000)xx50="0.005 mol = 0.005 g eq"` `"0.005 g eq. of NaOH neutralize 0.005 g eq. of "H_(2)SO_(4)," therefore, "H_(2)SO_(4)" left unneutralized in the solution"` `=0.02-0.005="0.015 g eq."` `"Total volume of solution after mixing = 150 mL = 0.150 L"` `therefore"Normality of "H_(2)SO_(4)" in the solution"=("0.015g eq")/(0.150L)="0.1 N. Hence, molairy"=("Normality")/("Basicity")=(0.1)/(2)="0.05 M"` (iv) `"50 mL of 0.1 N "H_(2)SO_(4)" contain "H_(2)SO_(4)=(0.1)/(1000)xx"50 g eq. = 0.005 g eq."` `"100 mL of 0.1 N NaOH contain NaOH"=(0.1)/(1000)xx"100 g eq. = 0.01 g eq."` `"0.005 g eq. of "H_(2)SO_(4)" neutralize 0.005 g eq. of NaOH, therefore, NaOHleft in the solution"` `=0.01-0.005="0.005 geq."` `"Total volume of solution after mixing = 150 mL = 0.150 L"` `therefore"Normality of NaOH in the solution"=("0.005 g eq")/("0.150 L")="0.033 N. Hence,"` `"molarity x "=("Mormality")/("Acidity")=(0.033)/(1)=0.033M` |
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