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Find the pH of the following soluitons: ...

Find the pH of the following soluitons:
(i) 3.2 g of hydrogen chloride dissolved in 1.0 L of water
(ii) 0.28 g of potassium hydroxide dissolved in 1.0 L of water .

Text Solution

Verified by Experts

In both the problems , the volume of the solution is the same as the volume of water
(i) Mass of hydrogen chloride =3.2 g
Molar mass of hydrogen chloride = 36.5 g `mol^(-1)`
No. of moles of hydrogen chloride `=((3.2 g))/((36.5 gmol^(-1)) = 0.0877 mol`
Volume of solution = 1.0 L
Molar concentration of solution`=((0.0877 mol))/((1.0L)) =0.0877 mol L^(-1) =0.0877 M`
The acid is strong and is completely ionised in solution as :
`HCloverset(aq)(to) underset(0.0877M)(H_(3)O^(+))+Cl^(-)(aq)`
`[H_(3)O^(+)] = 0.0877 M`
`pH=- log [H_(3)O^(+)] =- log (0.0877)=- log (8.77 xx 10^(-2))`
`=(2-log 8.77) =(2-0.94) = 1.06`
(ii) Mass of potassium hydroxide (KOH) = 0.28 g
Molar mass of KOH = 56. 0 g `mol^(-1)`
`:. [OH^(-)]` in the resulting solution `=0.1 xx 10^(-2) xx (1000)/(500) =2.0xx10^(-3) M`
`[H_(3)O^(+)]=(K_(w))/[[OH^(-)]]=((10^(-14)M^(2)))/((2.0xx10^(-3)M))=5xx10^(-12)M`
`pH=-log [H_(3)O^(+)] =- log (5.0 xx 10^(-12))`
`=(12 - log 5) =(12-0.69897)= 11.30103`
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