# Using the following reaction: C10H15ON(aq) +H20(l) = C10H15ONH+ (aq) + OH- (aq), a 0.035M solution of ephedrine has a pH of 11.33, how would you calculate the equilibrium concentrations of C10H15ON(aq), C10H15ONH+ (aq),and OH- (aq)?

Jan 9, 2016

Here's what I got.

#### Explanation:

Your strategy here will be to use an ICE table to help you find the equilibrium concentrations for two of the three chemical species that are taking part in the reaction.

It's important to notice that the problem provides you, albeit indirectly, with the equilibrium concentration of $\left[{\text{OH}}^{-}\right]$.

As you know, a solution's pH and its pOH have the following relationship at room temperature

$\textcolor{b l u e}{\text{pH" + "pOH} = 14}$

As you know,

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

Combine these two equations to get

overbrace(14 - "pH")^(color(red)("pOH")) = - log( ["OH"^(-)])

This means that you have

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

This is equivalent to

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

["OH"^(-)] = 10^(-(14 - 11.33)) = 10^-2.67 = 2.14 * 10^(-3)"M"

Now, ephedrine, a weak organic base, will not dissociate completely in aqueous solution. Instead, an equilibrium will be established between the unionized base and the ions it produces in solution.

An ICE table for this equilibrium would look like this

${\text{ " "C"_10"H"_15"ON"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons "C"_10"H"_15"ONH"_text((aq])^(+) " "+" " "OH}}_{\textrm{\left(a q\right]}}^{-}$

color(purple)("I")" " " "0.035" " " " " " " " " " " " " " " " " " " "0" " " " " " " " " " " " " " " " "0
color(purple)("C")" " " "(-x)" " " " " " " " " " " " " " " " " "(+x)" " " " " " " " " " "(+x)
color(purple)("E")" " " "0.035-x" " " " " " " " " " " " " " " " " "x" " " " " " " " " " " " " "x

Now, you know that the equilibrium concentration of the hydroxide ions is equal to $2.14 \cdot {10}^{- 3} \text{M" = "0.00214 M}$. As you can see from the ICE table, this equilibrium concentration is equal to $x$.

This of course tells you that

$\left[\text{C"_10"H"_15"ONH"^(+)] = ["OH"^(-)] = x = color(green)("0.00214 M}\right)$

The equilibrium concentration of the unionized base will be

["C"_10"H"_15"ON"] = "0.035 M" - "0.00214 M"

["C"_10"H"_15"ON"] = "0.0329 M"

I'll leave the values rounded to three sig figs.