Question

Calculate pH for 0.10 mole NH3 dissolved in 2L of 0.050 M NH4NO3?

Answer 1

`"pH" = 9.26`

Explanation:

I will assume that you're not familiar with the Henderson - Hasselbalch equation, which allows you to calculate the pH or pOH of a buffer solution.

So, you're interested in finding the pH of a solution that contains `0.10` moles of ammonia, `"NH"_3`, dissolved in `"2 L"` of `"0.050 M"` ammonium nitrate, `"NH"_4"NO"_3` solution.

As you know, ammonia is a weak base, which of course means that id does not ionize completely in aqueous solution to form ammonium ions, `"NH"_4^(+)`, its conjugate acid, and hydroxide ions, `"OH"^(-)`.

Instead, the following equilibrium will be established

`"NH"_text(3(aq]) + "H"_2"O"_text((l]) rightleftharpoons "NH"_text(4(aq])^(+) + "OH"_text((aq])^(-)`

Now, you're dissolving the ammonia in a solution that already contains ammonium ions, since ammonium nitrate, a soluble ionic compound, dissociates completely to form

`"NH"_4"NO"_text(3(aq]) -> "NH"_text(4(aq])^(+) + "NO"_text(3(aq])^(-)`

Notice that ammonium nitrate dissociates in a `1:1` mole ratio with the ammonium ions, which means that

`["NH"_4^(+)] = ["NH"_4"NO"_3] = "0.050 M"`

The concentration of ammonia in this solution will be

`color(blue)(c = n/V)`

`["NH"_3] = "0.10 moles"/"2 L" = "0.050 M"`

Use an ICE table to determine the equilibrium concentration of hydroxide ions in this solution

`" ""NH"_text(3(aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "NH"_text(4(aq])^(+) " "+" " "OH"_text((aq])^(-)`

`color(purple)("I")" " " "0.050" " " " " " " " " " " " " " " " "0.050" " " " " " " " "0`
`color(purple)("C")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " "(+x)`
`color(purple)("E")" "(0.050-x)" " " " " " " " " " " "0.050+x" " " " " " "x`

By definition, the base dissociation constant, `K_b`, will be equal to

`K_b = (["NH"_4^(+)] * ["OH"^(-)])/(["NH"_3])`

`K_b = ( (0.050+x) * x)/(0.050-x)`

You can find the value for `K_b` here

http://www.bpc.edu/mathscience/chemistry/table_of_weak_bases.html

So, you have

`1.8 * 10^(-5) = ( (0.050+x) * x)/(0.050-x)`

Rearrange this equation to get

`x^2 + (0.050 + 1.8 * 10^(-5)) * x - 9 * 10^(-7) = 0`

This quadratic solution will produce two solutions for `x`, one positive and one negative. Since `x` represents concentration, you can discard the negative value.

You will thus have

`x = 1.799 * 10^(-5)`

This means that you have

`["OH"^(-)] = x = 1.799 * 10^(-5)"M"`

The pOH of the solution will be

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

`"pOH" = - log(1.799 * 10^(-5)) = 4.74`

The pH of the solution will thus be

`color(blue)("pH" = 14 - "pOH")`

`"pH" = 14 - 4.74 = color(green)(9.26)`

SIDE NOTE It is important to notice here that when you have equal concentrations of weak base and conjugate acid, the pOH of the solution will be equal to the `pK_b` of the weak base

`color(blue)(pK_b = - log(K_b))`

`pK_b = -log(1.8 * 10^(-5)) = 4.74`

The Henderson - Hasselbalch equation for weak base / conjugate acid buffers looks like this

`color(blue)("pOH" = pK_b + log( (["conjugate acid"])/(["weak base"])))`

Notice that when

`["conjugate acid"] = ["weak base"]`

you have

`"pOH" = pK_b + log(1) = pK_b`