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)
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.
To determine the pH of a solution resulting from the reaction between an acid (HNO3) and a base (NaOH), you need to consider the stoichiometry of the reaction. The balanced equation is as follows:
HNO3 + NaOH → NaNO3 + H2O
From the given volumes and concentrations, you correctly calculated the number of moles of NaOH and HNO3 present:
[NaOH] = 0.025 L * 0.400 M = 0.010 mol
[HNO3] = 0.050 L * 0.250 M = 0.0125 mol
To determine the excess or limiting reactant, you correctly deducted the moles of NaOH from the moles of HNO3:
[HNO3] - [NaOH] = 0.0125 - 0.010 = 0.0025 mol
However, the next step, where you calculated [H3O] is incorrect. The number of moles of HNO3 used should be divided by the total volume of the resulting solution, not just the sum of the initial volumes of the reactants:
Total volume = 0.025 L + 0.050 L = 0.075 L
Correct calculation:
[H3O] = 0.0025 mol / 0.075 L = 0.0333 M
Finally, to find the pH, take the negative logarithm (base 10) of [H3O] using the equation:
pH = -log[H3O]
Therefore, pH = -log(0.0333) ≈ 1.48
Please note that this is the corrected calculation based on the information provided. However, if your professor still deems it incorrect, there might be additional factors or information that need to be considered. It is recommended to consult with your professor for further clarification.