Calculate pH of:
(a) `10^(-3)NHNO_(3)`, (b) `10^(-3)MH_(2)SO_(4)`,
(c ) `10^(-3)NH_(2)SO_(4)`, (d) `0.01N HCI`,
(e ) `10^(-8)NHCI`, (f) `10^(2)M HCI`.

Explanation

Strong acids ionise completely at normaly dilutions.
(a) `{:(10^(-3)HNO_(3):, HNO_(3)rarr,H^(+)+,NO_(3)^(-)), ("Conc.before ionisation", 10^(-3)N,0,0), ("Conc. after ionisation", 0,10^(-3),10^(-3)):}`
`:. [H^(+)]=10^(-3)"mol/litre or eq /litre"`
`(H^(+) is "monovalent")`
`:. pH= -log[H^(+)]= -log 10^(-3)`
`:. pH=3`
(b) `{:(10^(-3)MH_(2)SO_(4):, H_(2)SO_(4)rarr,2H^(+)+,SO_(4)^(2-)), ("Conc.before ionisation", 10^(-3)M,0,0), ("Conc. after ionisation", 0,2xx10^(-3),10^(-3)):}`
Mole ratio of `H_(2)SO_(4):H^(+)+SO_(4)^(2-)::1:2:1`
`[H^(+)]= 2xx10^(-3)M`
`.: pH= -log[H^(+)]= -log 2xx10^(-3)`
`:. pH= 2.6989`
(c ) `{:(10^(-3)NH_(2)SO_(4):, H_(2)SO_(4)rarr,2H^(+)+,SO_(4)^(2-)), ("Conc.before ionisation", 10^(-3)N,0,0), ("Conc. after ionisation", 0,10^(-3),10^(-3)):}`
`.: "Equal equivalent of a substance gives equal equilibrium of its components"`.
`:. [H^(+)]= 10^(-3)M`
`pH= -log [H^(+)]`
`:. pH=3`
(d) `{:(0.01NHCI:,HCIrarr,H^(+)+,CI^(-)), ("Conc.before ionisation", 10^(-3)N,0,0), ("Conc. after ionisation", 0,10^(-2),10^(-2)):}`
`[H^(+)]=10^(-2)M`
`:. pH= -log [H^(+)]`
`:. pH=2`
(e ) `10^(-8)N HCI:`
Sol. I: `{:(,HCIrarr,H^(+)+,CI^(-)), ("Conc.before ionisation", 10^(-8)N,0,0), ("Conc. after ionisation", 0,10^(-8),10^(-8)):}`
`:. [H^(+)]=10^(-8M) but pH=8` is not possible because it is acid. Now `[H^(+)]=10^(-7)M` are alerady present in solution and since `10^(-8)lt10^(-7)` and thus, it should not be neglected.
`.: [H^(+)]=10^(-8)+10^(-7)`
`= 1.1xx10^(-7)M`
`:. pH= 6.9586`
Sol. II.
The above laks with discrepancy that dissiation of `H_(2)O`, a weak electrolyte is also suppressed in pressence of HCI due to common ion effect and thus, `[H^(+)]_(H_(2)O)` but will be lesser than `10^(-7)`. Therefore, dissociation of `H_(2)O` in presence of `10^(-8)H_(2)^(+)`.
`H_(2)OhArrunderset((10^(-8)+a))(H^(+))+underset(a)(H^(+))`
`:. K_(w)= (10^(-8)+a)a`
`a=0.95xx10^(-7)`
`:. [H^(+)]=10^(-8)+0.95xx10^(-7)`
`= 1.05xx10^(-7)`
`:. pH= 6.9788`
(f) `{:(10^(2)MHCI:,HCIrarr,H^(+)+,CI^(-)), ("Conc.before ionisation", 10^(-2)M,0,0), ("Conc. after ionisation", 0,10^(-2),10^(-2)):}`
`:. [H^(+)]=10^(2)M`
`:. pH= -2`
But this is not true. This may be explained as follows: Sorenson's originally intended pH to be related to `[H^(+)]`, but his fundamental method of mesurement method of mesurement - the hydrogen electrode - is now know to depend on thermodynamics activity and thus, `a_(H^(+))= [H^(+)]`. Thus, pH defined by - log `[H^(+)]` is not only of little theroreactical significance, but in fact cannot be mesured directly. It has therefore, came to be accept that `pH= -log a_(H^(+))`, i.e., pH of `10^(2)MHCI` cannot be calculated and it practically lies near to zero.