What is the pH of the solution when 50.00 mL of 0.250 M HNO3, nitric acid, has been treated with 25.00 mL of 0.400 M NaOH?
Below is the work I performed, however, my professor deemed the [H3O] as incorrect. Where did I go wrong?
[NaOH] = .025 L x 0.400 M = .010 mol
[HNO3] = .050 L x 0.250 M = .0125 mol
[HNO3] > [NaOH] so: .0125-.010 = .0025
.025+.050 = .075 L
[H3O] = .0025/.075 = .033
pH = -log(.033)
1 answer
Alex, you got me. I don't know. It looks ok to me. USUALLY it is a matter of an incorrect number of significant figures but this looks ok to me. The problem asks for pH and you didn't finish the calculation. Perhaps that is the problem.
3 answers