A nitric acid solution is titrated with a standardized 0.0206 M. NaOH (aq) solution. If 17.86 mL of the NaOH solution was required to neutralize 25.00 mL of HNO3 (aq) what was the pH of the original nitric acid solution?
Do i start off by doing 17.86 * 0.0206/ (17.86 + 25) Where do i go from here?? i keep getting stuck.. thank you so much if any one can help even somewhat!
2007-03-31 06:20:29 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
0.01786 L x 0.0206M = 0.0003679mol NaOH
1molNaOH for 1 mol of nitric acid so there are 0.0003679 moles of nitric acid in 25ml.
C = n/v = 0.01472M HNO3
Ph = -log[H] since HNO3 -> H+ + NO3 - is a strong acide (i belive) than 1 mol of HNO3 = 1 mol of H+
That makes Ph = -log[0.01472MH+]
1.832 pH
2007-03-31 06:30:24 · answer #1 · answered by w1ckeds1ck312121 3 · 0⤊ 0⤋
Are you asking for A Mathematical Solution or a Practical One for the Actual Experiment?
Thanks, RR
2007-03-31 06:23:49 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Yes.
2007-03-31 06:46:06 · answer #3 · answered by ag_iitkgp 7 · 0⤊ 0⤋