What is the change in ph after addition of 10.0 ml of 1.0 m sodium hydroxide to 90.0 ml of a 1.0 m nh3/1.0 m nh4+ buffer? [kb for ammonia is 1.8 x 10-5]?
I assume you mean M(molar) and not m(molal).
First, calculate the pH of the solution before any NaOH is added. Use the Henderson-Hasselbalch equation.
pH = pKa + log (NH3)/(NH4^+)
Then calculate the pH after the addition of the NaOH. I would use millimols for this because it makes it easier than using concentration. Technically, you're supposed to use concentration but the volume of the solution cancels so the answer using mmols and the answer using concentration is the same.
mmols NH3 = mL x M = 90
mmols NH4^+ = mL x M = 90
mmols NaOH added = mL x M = 10
.....NH4^+ + NaOH ==> NH3 + H2O + Na^+
I....90.......0.......90..........
add...........10...............
C...-10......-10......+10
E....80.......0........100
Plug the E line into the HH equation and solve for pH.
Subtract the two pH values to find the difference.
To calculate the change in pH after adding sodium hydroxide (NaOH) to the ammonia/nh4+ buffer solution, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the ratio of the concentrations of the weak acid (NH4+) and its conjugate base (NH3).
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
In this case, NH4+ acts as the weak acid (HA) and NH3 acts as its conjugate base (A-).
First, we need to calculate the initial concentrations of NH4+ and NH3. Since we are given that the 1.0 M NH3/1.0 M NH4+ buffer, the initial concentration of both NH4+ and NH3 is 1.0 M.
Next, we need to determine the pKa of the ammonium ion (NH4+). Since NH4+ is the conjugate acid of NH3, we can use the Kb value for NH3 to calculate the pKa using the following relationship:
pKa + pKb = 14
Given that the Kb for NH3 is 1.8 x 10^-5, we can rearrange the equation to solve for pKa:
pKa = 14 - pKb
pKa = 14 - (-log10(1.8 x 10^-5))
pKa = 14 - (-4.74)
pKa = 18.74
Now, we can substitute the values into the Henderson-Hasselbalch equation to calculate the initial pH:
pH = 18.74 + log(1/1)
pH = 18.74
Since adding NaOH will react with NH4+ to form more NH3, we need to calculate the final concentrations of NH4+ and NH3 after the addition of NaOH. To do this, we need to consider the stoichiometry of the reaction between NH4+ and NaOH. Since NaOH is a strong base, it will react completely with NH4+ in a 1:1 ratio. Therefore, the amount of NH4+ consumed by the addition of 10.0 mL of 1.0 M NaOH is:
Moles of NH4+ consumed = (10.0 mL) * (1.0 M) = 10.0 mmol
Since the initial concentration of NH4+ is 1.0 M, the final concentration of NH4+ after the addition of NaOH is:
Final concentration of NH4+ = (Initial concentration of NH4+) - (Moles of NH4+ consumed) / (Total volume of the solution)
Final concentration of NH4+ = (1.0 M) - (10.0 mmol) / (100.0 mL)
Final concentration of NH4+ = 0.9 M
Since there is no new NH3 formed, the concentration of NH3 remains the same.
Now, we can use this new NH4+ concentration and the concentration of NH3 (1.0 M) in the Henderson-Hasselbalch equation to calculate the final pH:
pH = 18.74 + log(1.0 / 0.9)
pH = 18.74 + log(1.11)
pH ≈ 19.14
Therefore, the change in pH after adding 10.0 mL of 1.0 M NaOH to the 90.0 mL of 1.0 M NH3/1.0 M NH4+ buffer is approximately 19.14 - 18.74 = 0.40.