Here's what I got.
One way to tackle this problem would be to use the fact that an aqueous solution at
#color(blue)(ul(color(black)("pH + pOH = 14")))#
#color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))#
to express the
#"pH" = 14 - [-log(["OH"^(-)])]#
#"pH" = 14 + log(["OH"^(-)])#
Now, you know that the concentration of hydroxide anions is given by the number of moles of hydroxide anions present for every
Use the molar mass of sodium hydroxide to convert the mass of sodium hydroxide to moles.
#90 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(39.997color(red)(cancel(color(black)("g")))) = "2.25 moles NaOH"#
Sodium hydroxide dissociates in a
Assuming that the volume of the solution does not change after you dissolve the sample of sodium hydroxide, you can say that the resulting solution will contain
This is, of course, equivalent to a molarity of
#["OH"^(-)] = "2.25 mol L"^(-1)#
Plug this into the equation to find the
#"pH" = 14 + log(2.25) = color(darkgreen)(ul(color(black)(14.4)))#
The answer is rounded to one decimal place, the number of sig figs you have for your values.
SIDE NOTE Keep in mind that this is only an approximation of the actual
In general, you can use the concentration of the hydronium cations to calculate the
The same can be said about the concentration of the hydroxide anions--you can use their concentration to find the
For more concentrated solutions, you need to use the activity of the hydronium cations, not their concentrations*.
#"pH" = - log(a_ ("H"_ 3"O"^(+)))#
#"pOH" = - log(a_("OH"^(-)))#