What is the [OH^-][OH] in a solution that has a [H_3O]=1.0*10^-6[H3O]=1.0106 MM?

1 Answer
Jul 30, 2016

The solution has a [OH^(-)][OH] of 1.0xx10^(-8)M1.0×108M

Explanation:

The answer can be obtained one of two ways:

  1. color(red)("Use the auto-ionization constant of water equation, Kw:"Use the auto-ionization constant of water equation, Kw:
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    All you would have to do is rearrange the equation to solve for [OH^(-)][OH] by dividing both sides of the equation by [H_3O^(+)]:[H3O+]:

[OH^(-)] =(1.0xx10^(-14))/[[H_3O^(+)]][OH]=1.0×1014[H3O+]

Plug in the known concentration of H_3O^(+)H3O+ ions:
[OH^(-)] =(1.0xx10^(-14))/(1.0xx10^(-6))[OH]=1.0×10141.0×106

[OH^(-)] = 1.0xx10^(-8)M[OH]=1.0×108M

2.
Determine the pH of the solution by taking the negative logarithm
(-log) of the concentration of H_3O^(+)H3O+ ions.

pH = -log(1.0xx10^(-6)) = 6pH=log(1.0×106)=6

Then obtain the pOH using the equation: pH + pOH = 14pH+pOH=14. Rearrange to solve for pOH:

pOH = 14 - 6 = 8pOH=146=8

Finally, you take the anti log (inverse of the natural logarithm), 10^("negative number")10negative number of the pOH to obtain the concentration of OH^(-)OH ions.

10^(-pOH)10pOH
10^(-8) = 1.0xx10^(-8)M108=1.0×108M