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Equal volumes of three acid solutions of `pH 3, 4` and `5` are mixed in a vessel. What will be the `H^(+)` ion concentration in the mixture? |
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Answer» For first solution, `pH =3 or - log [H^(+)] =3 or log [H^(+)] =- 3=overset(-)(3).0` ` [H^(+)] = " Antilog "[H^(+)] = " Antilog "(overset(-)(3).0) = 10^(-3)M` Similarly ,for second solution, " "`pH=4 and [H^(+)] =10^(-4) M` For third solution " "`pH =5 and [H^(+)] =10^(-5) M` Total `[H^(+)]` in the solution `=(10^(-3)+ 10^(-4) +10^(-5))=10^(-3) (1+10^(-1) +10^(-2))` `=10^(-3)(1+0.1 +0.01) = 1.11 xx 10^(-3)M` Since equal volumes of the three solutions have been mixed. `[H^(+)]` of the resulting solutin `=((1.11 xx 10^(-3)M))/(3) = 3.7 xx 10^(-4) M` |
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