Assuming complete dissociation, calculate the `pH` of the following solutions,
a. `0.003 M HCl, b. 0.005 M NaOH`,
c. `0.002 M HBr, d. 0.002 M KOH`

Explanation

(a) pH value of 0.003 M HCI solution.
`HCI(l)+H_(2)O(l) overset((aq))(to) H_(3)O^(+) +CI^(-)(aq)`
`0.003 M`
`pH =- log [H_(3)O^(+)] =- log (3xx 10^(-3))`
`=- (log 3-3 log 10) =(3 - log 3) =3 -0.477 =2.53`
(b) pH value of 0.005 M NaOH
`NaOH(s) overset((aq))(to) Na^(+) (aq) + underset(0.005 M)(OH^(-) (aq))`
`[OH^(-)] =0.005 M= 5 xx 10^(-3) M`
`[H_(3)O^(+)] =(K_(w))/[[OH^(-)]] =(((10^(-14) M^(2)))/((5xx10^(-3)M))) =2 xx 10^(-12) M`
`pH =- log [H_(3)O^(+)] =- (log 2 xx 10^(-12))`
`=- (log2 -12 log 10) =(12 -log 2) =(12 - 0.3010) =11.7`
(c) pH value of 0.002 M HBr
`HBr(l) +H_(2)O(l)overset((aq))(to) H_(3)O^(+) (aq) +Br^(-) (aq)`
`0.002 M`
`pH =- log [H_(3) O^(+)] =- log (2xx10^(-3)) =(3 - log 2) =3 - 0.3010 =2.7`
(d) pH value of 0.002 M KOH ltbr. `KOH (s) overset((aq))(to) K^(+) (aq) + underset(=0.002 M)(OH^(-)(aq))`
`[OH^(-)] =0.002 M =2 xx 10^(-3) M`
`[H_(3)O^(+)] =(K_(w))/[[OH^(-)]] = ((10^(-14) M^(2)))/((2xx10^(-3)M)) =5 xx 10^(-12) M`
`pH =- log [H_(3)O^(+)] =- log (5 xx 10^(-12))`
`=(12 - log 5) =(12 - 0.6989) = 11.30`