What is [OH‾] in an aqueous solution at 298 K in which [H+] = 5.35 *10^(3) M.
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To solve this you need to find the pH, then the pOH, and then you can solve for the [OH-]
to find pH, take -log[H+]
then to find pOH, take 14 - pH = pOH (pH + pOH = 14)
then take 10^-pOH
This will get you from [H+] to the [OH-]
Hope this helps!
to find pH, take -log[H+]
then to find pOH, take 14 - pH = pOH (pH + pOH = 14)
then take 10^-pOH
This will get you from [H+] to the [OH-]
Hope this helps!
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There is a formula that states the auto-ionization for pure water (Kw) is 1.0e-14. Since Kw=[OH-]x[H+] we can plug in for what you have: 1.0e-14/5.35e3 = 1.87e-18 = [OH-]