A buffer was prepared by mixing 1.00 mol of ammonia 𝑁𝐻3 𝐾𝑎 =

Question

A buffer was prepared by mixing 1.00 mol of ammonia 𝑁𝐻3 𝐾𝑎 =

5.8 × 10−10 and 1.00 mol of ammonium chloride 𝑁𝐻4𝐶𝑙 to form an
aqueous solution with a total volume of 1.00 liter. To 500 mL of this
solution was added 30.0 mL of 1.00 M 𝑁𝑎𝑂𝐻. What is the resulting
pH of the final solution?

Answer 1

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Answer 2

Thank u veryy helpfull

Answer 3

To find the resulting pH of the final solution, we need to understand the chemical reactions that occur in the solution.

Step 1: Write the balanced chemical equation for the reaction between ammonia (NH3) and ammonium chloride (NH4Cl) in water:

NH3 + NH4Cl ⇌ NH4+ + NH3OH-

In this equation, NH3 donates a proton (H+) to ammonium chloride to form ammonium ion (NH4+) and a hydroxide ion (OH-). This reaction helps in maintaining the pH of the solution.

Step 2: Calculate the concentration of NH3 and NH4+ after mixing the ammonia and ammonium chloride:

Initial concentration of NH3 = 1.00 mol/L
Initial concentration of NH4Cl = 1.00 mol/L

Total volume of the solution = 1.00 L

Concentration of NH3 and NH4+ in the final solution = (initial concentration x initial volume) / final volume
Concentration of NH3 and NH4+ in the final solution = (1.00 mol/L x 1.00 L) / 0.50 L = 2.00 mol/L

Step 3: Now, calculate the moles of NaOH added to the solution:

Volume of NaOH added = 30.0 mL = 0.030 L
Concentration of NaOH = 1.00 M

Moles of NaOH = concentration x volume = 1.00 M x 0.030 L = 0.030 mol

Step 4: Calculate the concentration of NH3OH- (the hydroxide ion from the reaction) in the final solution:

Concentration of NH3OH- = moles of NH3OH- / final volume
Concentration of NH3OH- = 0.030 mol / 0.530 L = 0.0566 M

Step 5: Determine the resulting pH using the concentration of NH3OH- (base) in the solution:

pOH = -log10(concentration of NH3OH-)
pOH = -log10(0.0566)
pOH = 1.24

pH + pOH = 14 (at 25°C)
pH = 14 - pOH = 14 - 1.24 = 12.76

Therefore, the resulting pH of the final solution is approximately 12.76.

Answer 4

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Answer 5

The initial buffer solution is 1L of 1 M NH3 + 1M NH4Cl and you take 500 mL of that solution and add NaOH to it. In the second solution you have 0.5 L consisting of 0.5 mol NH3 and 0.5 mol NH4Cl. You add 30 mL of 1 M NaOH which is 0.03 x 1 = 0.03 mol OH^-.

................NH4^+ + OH^- ==> NH3 + H2O
I...............0.5............0.................0.5....................
add........................0.03..................................
C...........-0.03.......-0.03.............+0.03
E...........0.47.............0................0.53
Plug the E line into the Henderson-Hasselbalch equation and solve for pH.
Post your work if you get stuck.