Question #107410
1) Calculate [H3O+] in the following aqueous solution at 25 ∘C: [OH−]= 1.0×10−9 M .
2) Calculate [H3O+] in the following aqueous solution at 25 ∘C: [OH−]= 2.6×10−2 M .
3) Calculate [H3O+] in the following aqueous solution at 25 ∘C: [OH−]= 7.0×10−12 M .
1
Expert's answer
2020-04-02T08:58:09-0400

The H3O+ ion is considered to be the same as the H+ ion as it is the H+ ion joined to a water molecule. The proton cannot exist in aqueous solution, due to its positive charge it is attracted to the electrons on water molecules and the symbol H3O+ is used to represent this transfer

[H+][OH]=1014[H^+][OH^-]=10^{-14}\\


1) [OH]=1.0×109M.[OH^−]= 1.0×10^{- 9} M .

[H+]=1014109=105M[H^+]=\frac{10^{-14}}{10^{-9}}=10^{-5 }M


2) [OH-] = 2.6×1022.6\times10^{-2}

[H+]=10142.6×102=3.84×1013M[H^+]=\frac{10^{-14} }{2.6\times10^{-2}}=3.84\times10^{-13}M


3) [OH]=[OH^-]=7×1012M7\times10^{-12}M

[H+]=10147×1012=1.43×103[H^+]=\frac{10^{-14}}{7\times10^{-12}}=1.43\times10^{-3}


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