Determine the pH of a 0.09 M KOH solution. Your answer must be within ± 0.1%?

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Answer 1 · 2018-05-04T04:14:26

#"pH" = 12.95#

#"KOH"# is a strong base, meaning that it dissociates completely. Therefore, the (balanced) equation is:
#"KOH (aq)" -> "K"^+ "(aq)" + "OH"^-##(aq)"#

This balanced equation tells us that every mole of #"KOH"# produces one mole of #"OH"^-#.
Therefore, we know that #["KOH"] = ["OH"^-] = 0.09 " M"#

Now, to find the #"pOH"#, we use the formula:
#"pOH" = -log["OH"^-]#

So:
#"pOH" = -log[0.09 " M"] ~~ 1.05#

However, the question wants the #"pH"#, not the #"pOH"#. To find this, we use:
#"pH" = 14 - "pOH"#

So:
#"pH" = 14 - 1.05 = 12.95#

The exact solution to #8# decimal places is #12.95424251#. This is within #0.1%#:

#|(12.95 - 12.95424251)/(12.95424251)| xx 100% ~~ 0.03%#

Hope this helps!