At 25°C, aqueous solutions with a pH of 6 have a hydroxide ion concentration, [OH-], of
pick one answer choice and explain
use the concept of Kw.
Kw=[H+][OH−]=1×10−14@25CKw = [H^+][OH^-] = 1×10^{-14} @ 25 CKw=[H+][OH−]=1×10−14@25C
If pH = 6, then[H+]=1×10−6mol/L,[H^+] = 1×10^{-6} mol/L,[H+]=1×10−6mol/L, so by
substitution 1×10−14=(1×10−6)[OH−]1×10^{-14} = (1×10^{-6})[OH^-]1×10−14=(1×10−6)[OH−]
[OH]=1×10−141×10−6=1×10−8mol/L[OH] = \frac{1×10^{-14}}{1×10^{-6}}= 1×10^{-8} mol/L[OH]=1×10−61×10−14=1×10−8mol/L
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