पूर्ण वियोजन मानते हुए निम्नलिखित विलियनों के pH ज्ञात कीजिए।
(क) 0.003 M HCl (ख) 0.005 MNaOH (ग)0.002 M HBr (घ) 0.002 M KOH

Explanation

(क) 0.03M HCI विलयन का pH मान
`HC(l) + H_2O (l) overset((aq))/(to) H_3O^(+) + Cl^(-) (aq)`
`pH = -log [H_3O^(+) ] = -log (3 xx 10^(-3))`
` = -(log3-3 log 10) = (3 - log 3)`
` = 3-0.477 = 2.53`
(ख) 0.005 M NaOH का pH मान
`NaOH (s) overset((aq))(to) Na^(+) (aq) + OH^(-) (aq)`
`[OH^-] = 0.005M = 5 xx 10^(-3) M`
` therefore [H_3O^+] = (K_w)/([OH^-]) = ((10^(-14) M^2))/((5 xx 10^(-3) M)) = 2 xx 10^(-12)M`
`pH = -log [H_3O^+] = -(log 2 xx 10^(-12))`
`= (-log 2 -12 log 10) = (12 - log 2)`
`= (12 - 0.3010 ) = 11.7`
(ग) 0.002 M HBr का pH मान
`HBr (l) + H_2O (l) overset((aq))/(to) H_3O^(+) (aq) + Br^(-) (aq)`
`pH = -log [H_3O^+] = -log (2 xx 10^(-3))`
` = (3 - log 2) = 3-0.3010 = 2.7`
(घ) 0.002 M KOH का pH मान
`KOH(s) overset((aq))/(to) K^(+) (aq) + OH^(-) (aq)`
` [OH^-] = 0.002 M = 2 xx 10^(-3) M`
`[H_3O^+] = (K_w)/([OH^-]) = ((10^(-14) M^2) )/((2 xx 10^(-3) M)) = 5 xx 10^(-12) M`
`pH = -log [H_3O^+] = -log (5 xx 10^(-12))`
` = -(log 5 - 12 log 10) = (12 - log 5)`
`= (12.0.6989) = 11.30`