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
Explanation:
Your buffer solution contains ammonia,
Sodium hydroxide,
The hydroxide anions will react with the ammonium cations to form ammonia and water
#"NH"_text(4(aq])^(+) + "OH"_text((aq])^(-) -> "NH"_text(3(aq]) + "H"_2"O"_text((l])#
Since you're dealing with a
This means that your initial solution contains
Now, notice that the ammonium cations and the hydroxide anions react in a
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
#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)|)))#