# Question #f84ab

##### 1 Answer

#### Explanation:

You pretty much lost me after you calculated the initial concentration of the weak base, so I'll solve the problem from scratch, that way you can follow along and see what went wrong.

You know that amphetamine is a weak base, which implies that an ionization equilibrium is established in aqueous solution

#"C"_ 9"H"_ 13"N"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "C"_ 9"H"_ 13"NH"_ ((aq))^(+) + "OH" _((aq))^(-)#

Now, the *base dissociation constant*,

#K_b = (["C"_ 9 "H"_ 13"NH"^(+)] * ["OH"^(-)])/(["C"_9"H"_13"N"])#

You also know that

#"p"K_b = - log(K_b)#

which implies that

#K_b = 10^(-"p"K_b)#

Now, the initial concentration of the base can be determined by using its **molar mass**

#200 color(red)(cancel(color(black)("mg")))/"L" * (1 color(red)(cancel(color(black)("g"))))/(10^3color(red)(cancel(color(black)("mg")))) * ("1 mole C"_9"H"_13"N")/(135.21 color(red)(cancel(color(black)("g")))) = 1.48 * 10^(-3)"M"#

If you take **equilibrium concentration** of amphetamine will be

#["C"_9"H"_13"N"] = ["C"_9"H"_13"N"]_0 - x#

#["C"_9"H"_13"N"] = 1.48 * 10^(-3) - x#

Plug this into the expression you have for the base dissociation constant to get

#10^(-"p"K_b) = (x * x)/(1.48 * 10^(-3) - x)#

#10^(-4.2) = x^2/(1.48 * 10^(-3) - x)#

#6.3 * 10^(-5) = x^2/(1.48 * 10^(-3) - x)#

Use the approximation

#1.48 * 10^(-3) - x ~~ 1.48 * 10^(-3)#

and rearrange the above equation to find the value of

#x^2 = 6.3 * 10^(-5) * 1.48 * 10^(-3)#

#x = sqrt( 6.3 * 10^(-5) * 1.48 * 10^(-3)) = 3.05 * 10^(-4)#

Since

#["OH"^(-)] = 3.05 * 10^(-4)"M"#

Consequently, you will have

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

which will get you

#"pOH" = - log(3.05 * 10^(-4)) = 3.52#

As you know, an aqueous solution at room temperature will have

#color(blue)(ul(color(black)("pH " + " pOH" = 14)))#

This means that the pH of the solution will be

#"pH" = 14 - 3.52 = 10.5#

The answer is rounded to on *decimal place*, the number of significant figures you have for the concentration of the base.

**However**, the problem here is that you **can't** really use the

#1.48 * 10^(-3) - x ~~ 1.48 * 10^(-3)#

approximation because

#(3.05 * 10^(-4))/(1.48 * 10^(-3)) xx 100% = 20.6% > color(red)(5%)#

The percent error this approximation would introduce is just *too significant* to ignore.

This means that you will have to solve this by way of a quadratic equation

#x^2 = 6.3 * 10^(-5) * (1.48 * 10^(-3) - x)#

This is equivalent to

#x^2 + 6.3 * 10^(-5)x - 9.324 * 10^(-8) = 0#

This quadratic equation will produce two solutions, one positive and one negative. Since *concentration*, the negative solution will not have any physical significance here.

Therefore, you will have

#x = 2.75 * 10^(-4)#

In turn, this will result in

#"pOH" = - log(2.75 * 10^(-4)) = 3.56#

Consequently, the pH of the solution will be

#color(darkgreen)(ul(color(black)("pH" = 14 - 3.56 = 10.4)))#

So, when rounded to one decimal place, the correct answer would be

#"pH" = 10.4" "color(darkgreen)(sqrt())" "# #" " color(red)(cancel(color(black)("pH" = 10.5)))#