what is the pH change after the addition of 10mL of 1.0M sodium hydroxide to 90mL of a 1.0M NH3/1.0M NH4^+ buffer? Kb ammonia= 1.8 x 10^-5

Question

what is the pH change after the addition of 10mL of 1.0M sodium hydroxide to 90mL of a 1.0M NH3/1.0M NH4^+ buffer? Kb ammonia= 1.8 x 10^-5

i found the pH for the original buffer to be 9.255, but i'm not sure how to go about the rest of the question, please, someone help me.

Answer 1

I would do this.

I assume you used the HH equation to arrive at the answer of 9.255 (with which I agree). At that time you substituted 1M each for base and acid and log 1 = 0 which makes pH = pKa. When you do that base/acid = 1 which means (base) = (acid) and the total is 1M.
90mL x 1M = 90 millimoles total or
45 mmols base and 45 mmoles acid.
The equation for the addition of NaOH is
............NH4+ + OH^- ==> NH3 + H2O
initial.....45.....0........45......0
add OH-............10...............
change.....-10....-10.....+10......+10
equil.......35.....0........55......10

pH = 9.255 + log(55/35) = ??
I would round the final answer to 2 places.

Answer 2

oh! i get it now, thank you very much =)

Answer 3

To determine the pH change after adding sodium hydroxide (NaOH) to the ammonia (NH3) and ammonium (NH4+) buffer solution, we need to consider the reaction that occurs between NaOH and NH4+.

First, let's calculate the concentration of NH4+ and NH3 in the original buffer solution:
Given:
Volume NH3/NH4+ = 90 mL
Concentration NH3/NH4+ = 1.0 M

We can assume that the solution is initially at equilibrium, so the concentration of NH4+ and NH3 is the same:
[ NH3 ] = [ NH4+ ] = 1.0 M

Next, we need to consider the reaction between NH4+ and OH- (from NaOH):
NH4+ + OH- ⇌ NH3 + H2O

Since NaOH is a strong base, it fully dissociates into Na+ and OH-. Therefore, the concentration of OH- added is equal to the concentration of NaOH added:
[ OH- ] = 1.0 M

Now, let's use the given Kb value to calculate the equilibrium constant expression for the reaction between NH4+ and OH-:
Kb = [ NH3 ][ OH- ] / [ NH4+ ]

Substituting the known values:
1.8 x 10^-5 = (1.0 M)(1.0 M) / [ NH4+ ]

Rearranging the equation:
[ NH4+ ] = (1.0 M)(1.0 M) / (1.8 x 10^-5)

Calculating:
[ NH4+ ] = 55.6 M

Since we know the initial concentration of NH4+ and OH- is 1.0 M, and assuming the reaction goes to completion, the concentration of NH4+ will decrease by 1.0 M, and the concentration of NH3 will increase by 1.0 M.

Now, let's calculate the concentration of NH4+ and NH3 after the reaction:
[ NH4+ ] = 55.6 M - 1.0 M = 54.6 M
[ NH3 ] = 1.0 M + 1.0 M = 2.0 M

To calculate the new pH, we can use the Henderson-Hasselbalch equation:
pH = pKa + log( [ NH3 ] / [ NH4+ ] )

Given that pKa for NH3/NH4+ is based on the Kb value, which can be calculated using the equation pKa + pKb = 14:
pKa = 14 - pKb
= 14 - (-log10(1.8 x 10^-5))

Calculating:
pKa = 9.920

Substituting the concentrations into the Henderson-Hasselbalch equation:
pH = 9.920 + log( 2.0 M / 54.6 M )

Calculating:
pH ≈ 9.659

Finally, to find the pH change, subtract the original pH from the new pH:
ΔpH = 9.659 - 9.255

Calculating:
ΔpH ≈ 0.404

Therefore, the pH changes by approximately 0.404 after the addition of 10 mL of 1.0 M sodium hydroxide to 90 mL of the 1.0 M NH3/1.0 M NH4+ buffer solution.