1.

Assuming complete dissociation , calculate the pH of the following solution: 0.003 M HCl (ii) 0.005 M NaOH (iii) 0.002 M HBr (iv) 0.002M KOH

Answer»

SOLUTION :(i) `HCl+aqtoH^(+)+Cl^(-) THEREFORE[H^+]=[HCl]=3 times10^-3M,pH=-log(3times10^-3)=2.52` <BR> (ii) `NaOH+aq to Na^(+)+OH^(-)`
`therefore [OH^+]=5 times 10^-3 M [H^+]=10^-14//(5 times 10^-3)=2 times 10^-12 M`
`pH=-log(2 times 10^-12)=11.70`
(iii) `HBr+aqtoH^(+)+Br, therefore[H^+]=2 times10^-3 M,pH=-log(2 times10^-3)=2.70`
(iv) `KOH+aq to K^(+) +OH^(-)`
`therefore [OH^+]=2 times 10^-3 M, [H^+]=10^-14//(2 times 10^-3)=5 times 10^-12`
`pH=-log(5 times 10^-12)=11.30`


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