Answer: The question has some details missing. here is the complete question ; An analytical chemist is titrating 88.4 mL of a 0.2700 M solution of ammonia (NH3 with a 0.4300 M solution of HNO3. The pK, of ammonia is 4.74 Calculate the pH of the base solution after the chemist has added 66.3 mL of the HNO3 solution to it . Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO3 solution
Explanation:
Given ;
- number of moles of base = 88.4 x 0.2700 = 23.868
- number of moles of acid = 0.4300 x 66.3 = 28.509
- This was after the equivalence point, as such net moles of acid = 28.509 - 23.868 = 4.641mol
- total volume of solution = 88.4 + 66.3 = 154.7mL
- Concentration of Acid = moles/volume = 4.641/154.7 = 0.03M
- From pH = -log[H^+] = -Log[0.03]
The Answer is C
<span> It can be separated by physical means and is uniform in composition.</span>
Ice, freezing, or melting then liquid evaporation or condensation, the vapor, deposition the ice again
