The hydrogen ion concentration of a 10^(-8)MHCl aqueous solution at 29

1.	The hydrogen ion concentration of a 10^(-8)MHCl aqueous solution at 298 K (K_(w) = 10^(-14)) is
Answer» `9.525xx10^(-8)M` `1.0xx10^(-8) M` `1.0xx10^(-6) M` `1.0525 xx 10^(-7)M` Solution :From acid, `[H^(+)]=10^(-8)M` From `H_(2)O][H^(+)]=[OH^(-)]=x "mol" L^(-1)` (say) `([H^(+)] or [OH^(-)]!=10^(-7)` M in presence of acid ) Thus, in the solution, `[H^(+)]=(10^(-8)+x) M and [OH^(-)]=XM` 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)pm sqrt(10^(-16)+4xx10^(-14)))/(2)` `=9.51xx10^(-8), "i.e.," [OH^(-)]=9.51xx10^(-8)`. Hence, `[H^(+)]=(10^(-14))/((9.51xx10^(-8)))=1.05xx10^(-7)M` ALTERNATIVELY, directly, the only CONCENTRATION which will give pH close to 7 but less than7 is (d).

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