i need help with finding the ph of a buffer solution initially consisting of 0.0400 moles NH3 and 0.025 moles NH4^+ and 20 ml 0.75 NaOH.
ka of NH+4 = 5.6 ×10^-10
Use the Henderson-Hasselbalch equation.
Construct an ICE chart. Add 20 mL of 0.75M (I guess that's M but you omitted the symbol) NaOH = 15 millimoles NaOH = 0.015 mols.
(Note: I simply assume the NaOH was not present initially and it was added later. That doesn't change the problem.)
..........NH4^+ + OH^- ==> NH3
initial.0.025......0.....0.0400
add..............0.015................
change..-0.015..-0.015.....+0.015
equil...0.010.....0.........0.015
Plug these numbers into the HH equation and solve for pH.
To find the pH of a buffer solution consisting of NH3 and NH4+ ions, we need to consider the equilibrium reaction between NH3 (ammonia) and NH4+ (ammonium ion):
NH3 + H2O ⇌ NH4+ + OH-
1. Calculate the moles of NH3 and NH4+ ions initially present in the buffer solution:
Given:
- Moles of NH3 = 0.0400 moles
- Moles of NH4+ = 0.025 moles
2. Determine the concentration of NH3 and NH4+ in the solution by dividing the moles by the total volume (20 mL = 0.02 L):
- Concentration of NH3 (C(NH3)) = 0.0400 moles / 0.02 L = 2 M
- Concentration of NH4+ (C(NH4+)) = 0.025 moles / 0.02 L = 1.25 M
3. Calculate the moles of OH- ions added by the NaOH solution:
Given:
- Volume of NaOH solution = 20 mL = 0.02 L
- Concentration of NaOH (C(NaOH)) = 0.75 M
- Moles of OH- ions = Concentration of NaOH (C(NaOH)) × Volume of NaOH solution (V(NaOH))
= 0.75 M × 0.02 L
= 0.015 moles
4. Determine the change in moles of NH3 and NH4+ ions due to the reaction with OH- ions:
The reaction between NH3 and OH- ions results in the formation of NH4+ ions.
The moles of OH- ions will react with NH3, converting it to NH4+ ions.
- Moles of NH3 consumed = Moles of OH- ions = 0.015 moles
- The moles of NH4+ produced will also be equal to the moles of OH- ions, as per the balanced equation.
5. Calculate the final moles of NH3 and NH4+ ions:
- Moles of NH3 remaining = Initial moles of NH3 - Moles of NH3 consumed
= 0.0400 moles - 0.015 moles
= 0.0250 moles
- Moles of NH4+ = Initial moles of NH4+ + Moles of NH4+ produced
= 0.025 moles + 0.015 moles
= 0.0400 moles
6. Calculate the new concentrations of NH3 and NH4+:
- Concentration of NH3 (C(NH3)) = Moles of NH3 remaining / Volume of solution
= 0.0250 moles / 0.02 L
= 1.25 M
- Concentration of NH4+ (C(NH4+)) = Moles of NH4+ / Volume of solution
= 0.0400 moles / 0.02 L
= 2 M
7. Calculate the pOH of the solution using the concentration of OH- ions:
- pOH = -log10(OH- concentration)
= -log10(0.015 M)
8. Determine the pH of the buffer solution:
Since the solution is a buffer and we have calculated the pOH, we can use the equation:
pH + pOH = 14
- pH = 14 - pOH
Substitute the calculated pOH value into the equation to find the pH value of the buffer solution.
Remember to round the pH value to the appropriate decimal places based on significant figures or the level of accuracy required in your calculations.