Calculate the `pH` of the following solutions:
a `10^(-2) M HCI`
b `10^(-3) M H_(2)SO_(4)`
c `0.2 xx 10^(-2) M NaOH`
d `0.3 xx 10^(-3) M Ca (OH)_(2)`

Explanation

Since `HCI` is monobasic strong acid, and is completely ionised
a. `pH = - log (10^(-2)) = 2`
b. Since `H_(2)SO_(4)` is dibasic strong (`n` factor `= 2`). Therefore concentration of `H^(o+)` ions `= 2 xx 10^(-3)N`
`{:(,H_(2)SO_(4)rarr,2H^(o+)+,SO_(4)^(2-),),("Initial",10^(-3)M,0,0,),("Final",0,2xx10^(-3)M,10^(-3)M,):}`
`pH =- log (2xx 10^(-3)) =- log2 - log (10^(-3))`
`=- 0.3 =3 = 2.7`
c. First method:
Since `NaOH` is monoacidic strong base, so
`[overset(Θ)OH] = 0.2 xx 10^(-2)M = 2xx 10^(-3)`.
So, first calculate `pOH` of `NaOH` and then calculate `pH` accordingly.
`pOH =- log [overset(Θ)OH] =- log(2) - log (10^(-3))`
`=- 0.3 +3 = 2.7`
`pH = 14 - poH = 14 - 2.7 = 11.3`
Second method: First calculate `[H_(3)O^(o+)]` and then calculate `pH` accordingly.
`K_(w) = [H_(3)O^(o+)] [overset(Θ)OH]`
`:. [H_(3)O^(o+)] = (K_(w))/(overset(Θ)([OH])) = (10^(-14))/(2xx10^(-3))`
`= 0.5 xx 10^(-11) = 5 xx 10^(-12)`
`:. pH =- log 5 - log (10^(-12)) =- 0.7 +12 = 11.3`
d. Since `Ca(OH)_(2)` is diacidic strong base, thus
`[overset(Θ)OH] = 2xx 0.3 xx 10^(-3) N (n` factor `= 2`)
or
`{:(,Ca(OH)_(2)rarr,Ca^(2+)+,2oversetΘOH,),(Initial,0.3xx10^(-3)M,0,0,),("Final",0,0.3xx10^(-3),2xx0.3xx10^(-3)M,):}`
`:. pOH =- log (2xx 0.3 xx 10^(-3)N)`
`=- log (6xx 10^(-4)N)`
`=- log (3xx2) - log (10^(-4))`
`=- log 3 - log 2+4`
`=- 0.48 - 0.3 +4 = 3.22`
Therefore, `pH = 14 - 3.22 = 10.78`