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Calculate the pH of 10^(-8) M HCl solution. |
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Answer» SOLUTION : Note. At the first instance, itappears that as `[H^(+)]=10^(-8)`, therefore, pH should be 8. But pH cannot be 8but should be less than 7 because the solutionis acidic. The reason is that from `H_(2)O, [H^(+)]=106(-7)` M which cannot be neglected in comparison to `10^(-8)`M. The pH can be calculated as FOLLOWS : From, acid, `[H^(+)]=[OH^(-)]=x "mol" L^(-1) ` (say)(as in presence of acid, `[H^(+)]` from `H_(2)O ~=10^(-7)M`) THUS, in the solution, now we have `[H^(+)]=(10^(-8)+x)M and [OH^(-)]=x M` But `[H^(+)] [OH^(-)]=K_(W) = 10^(-14)` `:.(10^(-8) +x)(x) = 10^(-14) or x^(2) + 10^(-8) x - 10^(-14) = 0` `x=(-10^(-8)pmsqrt(10^(-16)+4xx10^(-14)))/(2) = (-10^(-8)+sqrt(401)xx10^(-8))/(2)` (Taking +ve value only) `=(19.025xx10^(-8))/(2) = 9.5125xx10^(-8)` i.e, `[OH^(-)]=9.5125xx10^(-8)M` `:. pOH = -log(9.5125xx10^(-8))=8-0.9783~=7.02 " " :. "" pH=14-7.02=6.98` Alternatively, in a simplified manner, we have From acid, `[H^(+)]=10^(-8)M, "From" H_(2)O , [H^(+)]=10^(-7)M` `:. "Total" [H^(+)]=10^(-8)+10^(-7) = 10^(-8) (1+10)=11xx10^(-8)M` `:. pH = -log [H^(+)]=-log(11xx10^(-8))=-[log 11 + log 10^(-8)]` `=-[1.0414-8]=6.9586~~ 6.96` |
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