Would an aqueous solution of 10^-9 M NaOH have a pH of 5?pOH = -log[OH^-] So, pOH = -log[1.0 x 10^-9] = 9 pH = 14.00 - 9 = 5 (acidic) Am I doing this right, or am I wrong?

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jerichorayel | College Teacher | (Level 2) Senior Educator

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If we are to consider the computation, the answer is correct. But If we try to look at the content of the solution, we cannot see a hydronium concentration [H+] = 1.0x10^-5.

If pH = 5, then [H+] = 10^-5 = 1.0x10^-5

We can also look at it this way: Consider we have a volume of water (pH=7) and then a specific amount of NaOH is added. The amount of the NaOH added is so small to dramatically change the pH of the solution. 

1.0x10^-9 M NaOH is so dilute to affect the pH of the solution. Thus we can conclude that the pH of the solution is closely neutral. 

Another answer can be derived with the comparison with the H+ and OH- concentration of water. At pH = 7, [H+] = [OH-] = 10^-7 = 1.0x10^-7M. 10^-9 is lower than 10^-7 to affect the pH intensely.  To get the pH of the solution therefore, we can add the NaOH concentration with the hydroxide concentration and then get the pOH and eventually the pH.

[OH-] = 10^-7 + 10^-9 = 1.01x10^-7

pOH = -log [1.01x10^-7]

pOH = 6.996

pH = 14 – pOH

pH = 14 – 6.996

pH = 7.004 ~ 7.00 neutral.

And lastly, remember that the solution contains basic solution right? Therefore it cannot be considered acidic (based on the computation that pH = 5) because what you only have is the concentration of the base.


Hope this helps.




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