A buffer solution is 1.30 M in #NH_3# and 1.20 M in #NH_4Cl#. If 0.120 moles of #NaOH# are added to 1.00 L of the buffer, what is its pH? Assume the volume remains constant. Kb of #NH_3 = 1.8 xx 10^(-5)#.

1 Answer
Mar 19, 2016

Answer:

#"pH" = 9.31#

Explanation:

Your buffer solution contains ammonia, #"NH"_3#, a weak base, and ammonium chloride, #"NH"_4"Cl"#, the salt of its conjugate acid, the ammonium cation, #"NH"_4^(+)#.

Sodium hydroxide, #"NaOH"#, si a strong base that dissociates completely in aqueous solution to form sodium cations, which are of no interest here, and hydroxide anions, #"OH"^(-)#.

The hydroxide anions will react with the ammonium cations to form ammonia and water #-># think neutralization reaction.

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

Since you're dealing with a #"1.0-L"# solution, you can treat molarity and number of moles interchangeably.

This means that your initial solution contains #1.30# moles of ammonia and #1.20# moles of ammonium cation (ammonium chloride dissociates in a #1:1# mole ratio to form ammonium cations and chloride anions).

Now, notice that the ammonium cations and the hydroxide anions react in a #1:1# mole ratio. This tells you that the reaction consumes equal numbers of moles of each reactant.

Moreover, for every mole of ammonium cations that reacts with one mole of hydroxide anions, one mole of ammonia is formed.

Since you're adding #0.120# moles of hydroxide buffer solution, you can say that the neutralization reaction will completely consume the added hydroxide anions and leave you with

#n_(OH^(-)) = 0 -># completely consumed

#n_(NH_4^(+)) = 1.30 - 0.120 = "1.18 moles NH"_4^(+)#

#n_(NH_3) = 1.20 + 0.120 = "1.32 moles NH"_3#

Since the volume of the buffer is assumed to be constant, you can say that the resulting solution will have

#["NH"_4^(+)] = "1.18 M" " "# and #" " ["NH"_3] = "1.32 M"#

Now, to calculate the pOH of the solution, use the Henderson - Hasselbalch equation

#color(blue)(|bar(ul(color(white)(a/a)"pOH" = pK_b + log( (["conjugate acid"])/(["weak base"]))|))#

Here you have

#color(purple)(|bar(ul(color(white)(a/a)color(black)(pK_b = - log(K_b))color(white)(a/a)|)))#

Plug in your values to find the pOH of the solution

#"pOH" = - log(1.8 * 10^(-5)) + log( (1.18 color(red)(cancel(color(black)("M"))))/(1.32color(red)(cancel(color(black)("M")))))#

#"pOH" = 4.74 + log(1.18/1.32) = 4.69#

To find the pH of the solution, use the equation

#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|))) -># true for aqueous solutions at room temperature

In your case, you will have

#"pH" = 14 - 4.69 = color(green)(|bar(ul(color(white)(a/a)9.31color(white)(a/a)|)))#