a.) Calculate [OH−] in the following aqueous solution at 25 ℃: [H3O+]= 1.9×10−8 M
b.) Calculate [OH−] in the following aqueous solution at 25 ℃: [H3O+]= 6.5×10−5 M .
c.) Calculate [OH−] in the following aqueous solution at 25 ℃: [H3O+]= 3.2×10−2 M .
we have a relation
[H3O+][OH-] = 1.0 x 10-14
[OH-] = 1.0 x 10-14 / [H3O+]
a) [OH-] = 1.0 x 10-14 / 1.9 x 10-8
[OH-] = 5.3 x 10-7 M
b) [OH-] = 1.0 x 10-14 / 6.5 x 10-5
[OH-] = 1.5 x 10-10 M
c) [OH-] = 1.0 x 10-14 / [3.2 x 10-2]
[OH-] = 3.1 x 10-13 M
a.) Calculate [OH−] in the following aqueous solution at 25 ∘C:[H3O+]= 1.9×10−8 Mb.) Calculate...
Calculate [H3O+] in the following aqueous solution at 25 ∘C: [OH−]= 1.0×10−9 M . Calculate [H3O+] in the following aqueous solution at 25 ∘C: [OH−]= 2.3×10−2 M . Calculate [H3O+] in the following aqueous solution at 25 ∘C: [OH−]= 6.1×10−12 M . Classify the solutions as acidic or basic. [OH−]=1.0×10−9 M[OH−]=1.0×10−9 M [OH−]=2.3×10−2 M[OH−]=2.3×10−2 M [OH−]=6.1×10−12 M[OH−]=6.1×10−12 M
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