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

Explanation

`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 low and 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`
`=5 times 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