# Question #0f970

##### 1 Answer

Here's what I got.

#### Explanation:

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")))#

and that

#color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))#

to express the *hydroxide anions*.

#"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 **mole ratio** to produce hydroxide anions, so you can say that you're adding **moles** of hydroxide anions to this solution.

*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 **moles** of hydroxide anions in

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 #"pH"# of the solution because calculating the #"pH"# of a solution using the concentration of the hydronium cations--or indirectly by using the concentration of the hydroxide anions--is only accurate for very dilute solutions.*

*In general, you can use the concentration of the hydronium cations to calculate the #"pH"# of a solution if this concentration does not exceed #"1 mol L"^(-1)#.*

*The same can be said about the concentration of the hydroxide anions--you can use their concentration to find the #"pOH"#, and thus the #"pH"#, of the solution if their concentration is #< "1 mol L"^(-1)#.*

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"^(-)))#

*Here* *is the activity of the hydronium cations and*

*is the*

**activity**of the hydroxide anions.