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Calculate the pH of the resultant mixtures : 10 mL 0.2 M Ca(OH)_2+ 25 mL 0.1 M HCl |
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Answer» Solution :MOLARITY =Mol/Litre, So, Mol=Molarity X Litre 10 mL 0.2 M `Ca(OH)_2=10/1000` L x 0.2 M `=2/1000 "mol" Ca(OH)_2` =2 millimole `Ca(OH)_2` 25 mL 0.1 M HCl = `25/1000` L x 0.1 M `=2.5/1000`mol HCl =2.5 millimole HCl REMAINING substrate of neutralization: `{:(Ca(OH)_2+, 2HCl to , CaCl_2+,2H_2O),("1 mol","2 mol","1 mol","2 mol"):}` Here, 2 mol HCl react with 1 mol `Ca(OH)_2` `therefore 2.5/1000` mol HCl `to 1.25/1000` mol `Ca(OH)_2` react. `therefore` The remaining mol of `Ca(OH)_2`after neutralisation. `(("Initial"),(2.0/1000))-(("After REACTION"),(1.25/1000))` `=0.75/1000` mol `Ca(OH)_2` is remainly Molarity of remaining `Ca(OH)_2=("mol"/"TOTAL volume (L)")` `=(0.75/1000)1000/35`=0.02143 M `[OH^-]=2[Ca(OH)_2]=2xx0.02143` =0.04286 M =`4.286xx10^(-2)` M pOH=-log `[OH^-]=-log (4.286xx10^(-2))` =-(-1.3678)=1.3679 pH=(14.0-pOH)=(14.0-1.3679) =12.632 =12.63 |
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