Calcutate `pH` of following solutions:
(i) 0.001 M `HNO_(3)`,(ii) 0.005m `H_(2)SO_(4)` ,(iii) 0.01 M KOH
(iv) `10^(-6)` M NaOH, (v)0.0008 M ` Ba(OH)_(2)`

Explanation

`(i) {:(0.001MHNO_(3),:,HNO_(3)rarr,H^(+),+,NO_(3)^(-)),(,t=0,10^(-3)M,0,,10^(-3)),(t=eq,,0,10^(-3),,10^(-3)):}`
` :. pH=-log10^(-3)=3.`
`(ii){:(0.001MH_(2)SO_(4),:,H_(2)SO_(4)rarr,SO_(4)^(2-),+,2H^(+)),(,t=0,0.005M,0,,0),(t=eq,,0,0.005,,10^(-2)):}`
` :.pH=log(10^(-2))=2.`
`(iii){:(0.001MKOH,:,KOHrarr,K^(+),+,OH^(-)),(,t=0,10^(-2)M,0,,0),(t=eq,,0,10^(-2),,2xx10^(-2)):}`
iv`{:(10^-8MNaOH,:,NaOHrarr,Na^(+),+,OH^(-)),(,t=0,10^(-8)M,0,,0),(t=eq,,0,10^(-8),,10^(-8)),(t=eq,,H_(2)Orarr,H^(+),OH^(-),),(,,,x,,x):}`
So, in solution `[OH^(-)]=(10^(-8)+x)M` and `[H^(+)]=x`
`rArr=x=9.5xx10^(-8)=[H^(+)] :. pH=7.022`
`(V){:(0.0008MBa(OH)_(2),:,Ba(OH)_(2)rarr,Ba^(2+),+,2OH^(-)),(,t=0,8xx10^(-4)M,0,,0),(t=eq,,0,8xx10^(-4),,2xx8xx10^(-4)):}`
`[OH^(-)]=16xx10^(-4)M`
`:. pOH=2.8`
`:. pH=11.2`