`10^(-5)M HCI` solution at `25^(@)C` is dilluted 1000 times. The pH of the diluted solution will

Options

A. be equal to 8
B. remain unchanged
C. lie between 5 and 6
D. lie between 6 and 7

Answer

D

Explanation

According to dilution equation,
`M_(i)V_(i)=M_(f)V_(f)`
`(10^(-5))(x)=M_(f)(1000 x)`
`M_(f)=10^(-8)`
Therefore, when `10^(-5)` M HCI is diluted 1000 times, the concentration of the resulting soluting will be `10^(-8)`M. Since HCI is a strong acid, the pH of the diluted solution is expected to be 8. This result cannot be accepeted because this value comes in the basic range. pH can be brought down in the acidic range if more `H^(+)` ions are considered in calculation. The other source of `H^(+)` ions is `H_(2)O.H^(+)` ion concentrations form the acid and water are comparable. Moreover, the concentration of `H^(+)` ions form `H_(2)O` is no longer `10^(-7)` M as the equilibrium of water is disturbed on the addition of HCI. Let us assume that when equilibrium is restableished, the concentration of `H^(+)` ions form `H_(2)O` is x mol `L^(-1)`
`2H_(2)O(aq.) hArr H^(+)underset(x)((aq.))+OH^(-)underset(x)((aq.))`
Total `H^(+)` concentration will be `C_(H^(+))` form `H_(2)O` and `C_(H^(+))` form HCI.
`K_(w)=C_(H^(+))C_(OH^(-))`
`1xx10^(-14)=(x+10^(-8))(x)`
`x^(2)+10^(-8)x-10^(-14)=0`
Solving the quardatic equation gives
`x=9.5xx10^(-8)`
Thus, total `C_(H^(+))=9.5xx10^(-8)+10^(-8)`
`= 1.05xx10^(-7) mol L^(-1)`
`pH= -(1ogC_(H^(+)))/(mol L^(-1))`
`= -1og (1.05xx10^(-7))`
`=6.98`
Now, remember the following guidellince while calculating the pH of an acid:
(i) `C_(H^(+)=C_(H^(+))` form acid only if `C_("acid")` is greater than `10^(-5) M`.
(ii) `C_(H^(+))=C_(H^(+))` form acid `+ C_(H^(+))` form water if `C_("acid")` lies between `10^(-8)` M and `10^(-5)` M.
(iii) `C_(H^(+))=C_(H^(+))` from water only if `C_("acid")` is far less than `10^(-8)` M.