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InterviewSolution
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What is the `pH` of the following solutions: a. `10^(-7)M NaOH` b. `10^(-8)M NaOH` c. `10^(2) M NaOH` |
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Answer» a. `pH` of `10^(-7) M NaOh` First method: The concentration of `overset(Θ)OH` ions from pure water is `10^(-7)M`. So add `[overset(Θ)OH]` from water and `[overset(Θ)OH]` from `NaOH` to get `[overset(Θ)OH]_("Total")` `:. [overset(Θ)OH]_("Total") = 10^(7)M +2xx10^(-7)M` `pOH =- "log" [overset(Θ)OH]_("Total")` `=- log (2xx10^(-7))` `=- log2 - log 10^(-7)` `=- 0.3011 +7 = 6.6989` `:. pH = 14 - 6.6989 = 7.3011` Second method: Due to common ion `(overset(Θ)OH)`, the supression of ionsiation of `H_(@)O` takes place in the presence of `NaOH`. Let `[H_(3)O^(o+)] = [overset(Θ)OH] = xM` `2H_(2)O(l)hArr underset(x M)(H_(3)O^(oplus))(aq)+ underset(x M)(overset(Theta)(O)H)(aq)...(i)` `NaOH rarr underset(10^(-7)M)(Na^(o+)(aq)) + underset(10^(-7)M)(overset(Θ)OH (aq)) ...(ii)` `[H_(3)O^(o+)]_("Total")` from equaitons (i) and (ii) `= (x +10^(-7))M`. At equilibrium `[H_(3)O^(o+)] [overset(Θ)OH] = 10^(-14) = Kw` `(x) (x +10^(-7)) = 10^(-4)` or `x^(2) +10^(-7) x - 10^(-14) = 0` `:. x = (-10^(-7)+sqrt((10^(-7))^(2)+(4xx10^(-14))))/(2)` `:. [H_(3)O^(o+)] = x = 0.616 xx 10^(-7)M` Therefore, `[overset(Θ)OH]_("total") = (10^(-7) +0.6.18 xx 10^(-7))M` `:. pOH =- log (1.618 xx 10^(-7)) = 6.7910` `pH = 14 - 6.7910 = 6.209` b. `pH` of `10^(-8)M NaOH` First method : `[overset(Θ)OH]_("total") = 10^(-8) ("from NaOH") +10^(-7) ("from" H_(2)O)` `= 10^(-7) (10^(-1) +1) = 1.1 xx 10^(-7)` `pOH =- log (1.1 xx 10^(-7)) =- log 1.1 - log 10^(-7)` `=- 0.0414 +7= 6.9586` `pH = 14 - 6.9586 = 7.0414`. Second method : Ket `x = [overset(Θ)OH] = [H_(3)O^(o+)]` from `H_(2)O` The `[overset(Θ)OH]_("Total")` is generated from the ionisation of NaOH dissociated and from ionisation of `H_(2)O` `:. K_(w) = (10^(-8)+x) (x) = 10^(-4)` or `x^(2) +10^(-8)x - 10^(-14) = 0` `:. x =(-10^(-8) +sqrt((10^(-8))^(2)+(4xx10^(-14))))/(2)` `[H_(3)O^(o+)] = x = 9.5 xx 10^(-8)` So, `pH = 7.02` c. pH of `10^(-2) M NaOh` `{:(,NaOHrarr,Na^(o+)+,overset(Θ)OH,),("Initial concentration",10^(2),0,0,),("Concentration after dissociation",0,10^(2),10^(2),):}` `pOH =- log (10^(2)) =-2` `pH = 14 - (-2) = 16` `pH gt 12`, only means that the `[overset(Θ)OH] gt 1M`. So pH scale becomes in between 2 and 16. |
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