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

Question

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

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.

Answer 2

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Answer 3

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Answer 4

Thank u veryy helpfull