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Calculate pH of 10^-8 M H_2SO_4 (ii) Calculate the concentration of hydrogen ion in moles per litre of a solution whose pH is 5.4. (iii) Calculate the pH of an aqueous solution obtained by mixing 50 ml of 0.2 HCl with 50 ml 0.1 M NaOH |
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Answer» Solution :`H_2SO_4 leftrightarrow^(H_2O)2H_3O^++SO_4^(2-)` `10^-8M 2 times10^-8 10^-8M` In this case the concentration of `H_2SO_4` is very lowand HENCE `[H_3O^+]` from water cannot be neglected. `therefore[H_3O^+]=2times10^-8` (from `H_2SO_4)+10^-7` (from water) `=10^-8(2+10)` `=12 times10^-8=1.2 times10^-7` `pH=-log_10[H_3O^+]` `=-log_10(1.2 times 10^-7)` `=7-log_10 1.2` `=7-0.0791=6.9209` (ii) pH of the solution=5.4 `[H_3O^+]`=antilog of (-pH) =antilog of (-5.4) =antilog of (-5+0.6)=6.6 `=3.981 times 10^-6` i.e, `3.98 times 10^-6 MOL dm^-3` (iii) No. of moles of HCl=`0.2 times 50 times 10^-3=10 times 10^-3` No. of moles of NaOH=`0.1 times 50 times 10^-3=5 times 10^-3` No. of moles of HCl after mixing=`10 times 10^-3 -5 times 10^-3` `=5times 10^-3` After mixing total volume =100 ML `therefore` COncentration of HCl in moles per litre `=(5 times 10^-3 MOLE)/(100 times 10^-3L)` `[H_3O^+]=5times10^-2M` `pH=-log(5times10^-2)` `=2-log5` `=2-0.6990` =1.30 |
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