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)#.

Question

Ellie Andrews · Accepted Answer

#"pH" = 9.31#

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.

Furthermore, one mole of ammonia is produced for every mole of ammonium cations that react with one mole of hydroxide anions.

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#

Given that the buffer's volume is presumed to be constant, you can state that the final solution will have

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

Now, apply the Henderson-Hasselbalch equation to determine the solution's pOH.

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

Here it is

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

Enter your values to determine the solution's pOH.

#"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#

Utilize the equation to determine the pH of the mixture.

#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 situation, you'll have

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

Naomi Aronson · Answer

undefined Calculate the new concentrations of NH₃ and NH₄⁺ after the addition of NaOH. Use the Henderson-Hasselbalch equation to find the new pH. The equation is pH = pKa + log([A-]/[HA]).

Olivia Anton · Answer

undefined To solve for the pH of the buffer solution after the addition of NaOH, we first need to determine how much of the NaOH reacts with the NH3 and NH4+ ions in the buffer solution. NH3 reacts with NaOH in a 1:1 ratio to form NH4+ and OH- ions. Therefore, the number of moles of NH3 that react with NaOH is equal to the number of moles of NaOH added.

After reacting, the NH4Cl in the buffer will contribute H+ ions through hydrolysis, which we need to consider for the pH calculation.

Then, we can calculate the concentrations of NH3, NH4+, and OH- ions after the reaction. Using the equilibrium constant Kb for NH3, we can calculate the concentration of OH- ions. Finally, we can use the concentrations of NH4+ and OH- ions to calculate the pH of the solution using the equation for the ionization of water.

After performing the calculations, we find the pH of the buffer solution after the addition of NaOH.