Calculate #pH# of #10^-9# #M# #NaOH#?

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Answer 1 · 2016-01-30T17:26:54

#pH=7#

Consider the two equilibria happening in solution:

#NaOHrightleftharpoonsNa^(+)+OH^-#

and

#H_2OrightleftharpoonsH^(+)+OH^-#

From #NaOH#: #[OH^-]=10^(-9)M#

From #H_2O#: #[OH^-]=10^(-7)M#

Since #[OH^-]_(H_2O)" >> "[OH^-]_(NaOH)#,

The #pOH# will be determined from the #[OH^-]_(H_2O)=10^(-7)M#

#pOH=-log([OH^-])=-log(10^(-7))=7#

Therefore, to calculate the #pH#:

#pH=pK_w-pOH=14-7=7#

Here is a video that explains in details this topic and treats a similar example:

Acids & Bases | pH of a Strong Acid & a Strong Base.