What is the `pH` of the following solutions:
a. `10^(-8)M HCI` b. `5 xx 10^(-8) M HCI`
c. `5xx10^(-10)M HCI` d. `10^(2) M HCI`

Explanation

`pH` of `10^(-8) M HCI`
First method:
`[H_(3)O^(o+)] = 10^(-8)` (from HCI) `= 10^(-7)` (from `H_(2)O)`
`= 10^(-7) (10^(-1) +1) = 1.1 xx 10^(-7)`
`pH =- log (1.1 xx 10^(-7))`
`=- log 1.1 - log 10^(-7)`
`=- 0.414 +7 = 6.9586`
Second method:
Let `x = [oversetΘ)OH] = [H_(3)O^(o+)]` from `H_(2)O`.
The `[H_(3)O^(o+)]_("total")` is generated from the ionisation of `HCI` dissolved and from ionisation of `H_(2)O`.
`:. K_(w) = (10^(-8)+x) (x) = 10^(-14)`
or`x^(2) +10^(-8)x - 10^(-14) = 0`
`:. x=(-10^(-8)+sqrt((10^(-8))_(2)+(4xx10^(-14))))/(2)`
`[overset(Θ)OH] = x = 9.5 xx 10^(-8)M`
So, `pOH = 7.02` and `pH = 6.98`
b. `pH` of `5xx10^(-8)M HCI` :
If the contribution of `[H_(3)O^(o+)]` from `HCI` is considered, i.e., `[H_(3)O^(o+)] = 5xx10^(-8)M`. Then `pH` would be `gt7`. This is not possible because `[H_(3)O^(o+)]` of nay acid solution, no matter how dilute it is, cannot be less than that of pure water alone. in such cases, the concentration of `[H_(3)O^(o+)]` made by water has to be taken.
Let `[H_(3)O^(o+)]` from water be `x M` in the presence of `5 xx 10^(-8) M HCI`.
`{:(,2H_(2)O,hArr,H_(3)O^(oplus),+,[overset(Theta)(OH)]),(,,,xM,,xM),(HCl+,H_(2)O,hArr,H_(3)O^(oplus),+,Cl^(Theta)),(,,,5xx10^(-8)M,,5 xx10^(-8)M):}`
At equilibrium, `[H_(3)O^(o+)] = (x +5 xx 10^(-8)), [overset(Θ)OH] = x`.
`K_(w) = [H_(3)O^(o+) [overset(Θ)OH] = (x +5 xx 10^(-5)) (x) = 10^(-14)`
`x = 0.78 xx 10^(-7) M`
`:. [H_(3)O^(o+)] = (5 xx 10^(-8) +0.78 xx 10^(-7))`
`= 1.28 xx 10^(-7)`
`pH =- log (1.28 xx 10^(-7)) =- (0.11-7) =6.89`
c. `pH` of `5 xx 10^(-10) M HCI`:
`HCI` is so diulte that its contribution to `[H_(3)O^(o+)]` is negligible as compared with the ionisation of water. Thus, `[H^(o+)] = 10^(-7)` and therefore, `pH = 7.00`
d. `pH` of `10^(-2)M HCI:`
`{:("Initial concentration",HCIrarr+H_(2)Orarr,H_(2)Orarr+,CI^(Θ),),("Concentration after",10^(2),0,0,),("dissociation",0,10^(2),10^(2),):}`
`pH =- log (10^(-2)) =- 2`
A nagative `pH` only means that the `[H^(o+)] gt 1M`. However, in actual practive, a negative `pH` is uncommon. Firstly, even strong acids (say `100% H_(2)SO_(4))` become partially dissociated ay hight concentration.
According to Sorenson, `pH` is related to thermodynamic activites rather than `[H^(o+)]`, i.e., on `a_(H)^(o+) = [H^(o+)] f_(H^(o+))`. In dilute solution activity coeffiecient, `f_(H^(o+))` is near enough to unity and thus, `a_(H^(o+)) = [H^(o+)]`. At high concentrations, the activity coefficient is less than unity. Thus, `pH` denfined by `-log [H^(o+)]`, which is not only of littele theoretical significance, but it in fact cannot be measured directly. Therefore, `pH` is redefined as :
`pH =- log_(10) a_(H^(o+))`
(This is what `apH` meter reading is a meausre of) i.e., `pH` of `10^(2) M HCI` cannot be calculated untiall `f_(H^(o+)` is known.
Nevertheless, there is mathematically no basis for not having a negative `pH`.