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Calculate the pH of each of the following solution (i) 100 ml of 0.1 M `CH_(3)COOH` mixed with 100 ml of 0.1 M NaOH. (ii) 100 ml of 0.1 M `CH_(3)COOH` mixed with 50 ml of 0.1 m NaOH (iii) `50 ml of 0.1 M CH_(3)COOH` mixed with 100 ml of 0.1 M NaOH. `K_(a)(CH_(3)COOH)=1.8xx10^(-5)` |
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Answer» (a) `underset("1 mole")(CH_(3)COOH)+underset("1 mole")(NaOH)toCH_(3)underset("1 mole")(COONa)+underset("1 mole")(H_(2)O)` m. moles of `Ch_(3)COOH=0.1xx100=10` m moles of `NaOH=100xx0.1=10` Therefore, m. moles of `CH_(3)COONa` is 10 Now in the solution only salt `CH_(3)COONa` is present, therefore, pH of solution can be calculated by the hydrolysis of `CH_(3)COONa` `CH_(3)COONa` is a salt of weak acid with strong base `pH=7+1/2pK_(a)+1/2logC` `=7+1/2xx4.75+1/2log((10)/(200))" "(pK_(a)=-log1.8xx10^(-5))` `=8.72" "pK_(a)=4.75` (b) `CH_(3)COOH+NaOHtoCH_(3)COONa+H_(2)O` `{:("Initial m moles",10,5,0,0),("m moles after reaction",10-5,0,5,5):}` Since in the solution, mixture of `CH_(3)COOH and CH_(3)COONa` is present therefore, solution is an acidic buffer and pH can be calculated as `pH=pK_(a)+log""(["salt"])/(["acid"])pH=4.75+log""(((5)/(200))/((5)/(200)))pH=4.75` (c ) `CH_(3)COOH+NaOHto CH_(3)COONa+H_(2)O` `{:("Initial m moles",5,10,0,0),("m moles after reaction",0,(10-5),5,5):}` Now in the solution strong base NaOH and salt of weak aid with strong base is present We can assume that `OH^(-)` ions concentration from the hydrolysis of salt is negligible because hydrlysis of `CH_(3)COONa` is suppressed by the presence of strong base NaOH. `underset(("from salt"))(CH_(3)COO^(-))+H_(2)OhArrCH_(3)COOH+OH^(-)` `therefore[OH^(-)]=([5])/([150])=((1)/(30))` `pOH=log[OH^(-)]=-log((1)/(30))=log30=1.4771` `pH+pOH=14` `pH=14-1.4771=12.52` |
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