Calculate pH of 2.0e-8M NaOH in Water

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To calculate the pH of a solution with 2.0e-8 M NaOH while considering the contribution of hydroxide ions (OH-) from water, the key equations involved are Kw = [OH-][H+] and the charge balance equation [OH-] = [H+] + [Na+]. The process requires rearranging the first equation to express [OH-] as Kw / [H+], which can then be substituted into the charge balance equation. This results in a quadratic equation that can be solved to find the pH, which is determined to be 7.04. For assistance with similar calculations, using tools like BATE for pH and acid-base equilibrium calculations is recommended.
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there are 2.0e-8 M NaOH in water. I have to calculate the pH of the solution considering the OH from the water. The answear to this is pH=7.04 I know this is calculated with the quadratic equation. what I don't get is how to set this up. I alway get a wrong answear. Can somone help me with this?
 
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gunnar said:
there are 2.0e-8 M NaOH in water. I have to calculate the pH of the solution considering the OH from the water. The answer to this is pH=7.04 I know this is calculated with the quadratic equation. what I don't get is how to set this up. I alway get a wrong answear. Can somone help me with this?

You have two equations:

Kw = [OH-][H+]

and charge balance:

[OH-] = [H+] + [Na+]

This is enough to solve the problem, but in case you don't know what to do next:

You have to get rid of [OH-], so rearrange first equation

[OH-] = Kw / [H+]

and use [OH-] in the second equation:

Kw/[H+] = [H+] + [Na+]

That's your quadratic equation.

Try BATE for such (and much more complicated) pH and acid - base equilibrium calculations.
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BATE - pH calculations, titration curves, hydrolisis
 
Worked out fine. Thanks a lot :smile:
 
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