पूर्ण वियोजन मानते हुए निम्नलिखित विलयनों के pH ज्ञात कीजिये-
(क) `0.003 M HCl`, (b) 0.005 M NaOH
( c) 0.002 M HBr, (d) 0.002 M KOH

Explanation

(a) 0.003 M HCl
`[H^(+)]=-log_(10)[H^(+)]=-log_(10)(3.0 xx 10^(-3))`
`=-log_(10)3 +3`
`=-0.477 + 3.0 = 2.523`
(b) 0.005 M NaOH
`[OH^(-)] = [NaOH] = 0.005 = 5.0 xx 10^(-3) M`
`[H^(+)] = K_(w)/([OH^(-)]) = (1.0 xx 10^(-14))/(5.0 xx 10^(-3)) = 2.0 xx 10^(-12)`
`therefore pH = -log_(10)[H^(+)]=-log_(10)(2.0 xx 10^(-12))`
`=-log_(10) 2+12`
`=-0.3010 + 12 = 11.699`
वैकल्पिक विधि
`pOH = -log_(10)[OH^(-)]= -log_(10)(5.0 xx 10^(-3))`
`=-log_(10) 5+3`
`=-0.699 + 3=2.301`
`therefore pH = 14-pOH = 14-2.301 = 11.699`
( c) `0.002 M HBr`
`pH = -log_(10)(5.0 xx 10^(-12))`
`-=-log_(10) (5+12)`
`=-0.6989 + 12 = 11.301`