Question #36e0b

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Answer 1 · 2017-02-15T00:06:36

#"pH" = 11#

The thing to remember about an aqueous solution at room temperature is that

#color(blue)(ul(color(black)(["OH"^(-)] * ["H"_ 3"O"^(+)] = 10^(-14))))#

Rearrange this equation to isolate the concentration of hydronium cations on one side

#["H"_ 3"O"^(+)] = (10^(-14))/(["OH"^(-)])#

Now, you should be aware that

#color(blue)(ul(color(black)("pH" = - log( ["H"_3"O"^(+)]))))#

Plug the expression you have for the concentration of the hydronium cations in this equation to get

#"pH" = - log( (10^(-14))/(["OH"^(-)]))#

In your case, you will have

#"pH" = -log( (10^(-14))/(10^(-3)))#

#"pH" = - log(10^(-11))#

#"pH" = -(-11) log(10)#

#color(darkgreen)(ul(color(black)("pH" = 11)))#