✔ 最佳答案
-initial ph of solution
[OH-] = 0.345 M
pOH = -log(0.345) = 0.462
pH = 14.00 - pOH = 13.538
-ph of solution if 12.00 ml of titrant was added
Moles OH- in original solution = 0.03000 L X 0.3450 M = 0.01035 mol OH-
Moles H+ in 12.00 mL titrant = 0.01200 L X 0.5000 mol/L = 6.000X10^-3 mol
Moles OH- remaining = 0.01035 - 6.00X10^-3 = 4.35X10^-3 mol
[OH-] remaining = 4.35X10^-3 mol / 0.04200 L = 0.1036 M OH-
pOH = 0.985, pH = 13.015
-volume of titrant needed to neuralize 30.00 ml of 0.3450 M of KOH
0.01035 mol NaOH will require 0.01035 mol HCl
Volume HCl = 0.01035 mol / 0.5000 mol/L = 0.0207 L = 20.70 mL HCl
-pH at equivalence pt
Since you are titrating a strong acid with a strong base, pH at equivalence point = 7.000
-pH of the solution if 25.00 ml of titrant was added to the volume of post equivalence point
By my reading of this, you will have added a total volume of HCl of 45.70 mL HCl (25 mL more than the 20.7 mL to reach equivalence point)
Moles HCl added = 0.04570 L X 0.500 mol/L = 0.02285 mol HCl added
Moles HCl unreacted = 0.02285 - 0.01035 = 0.01250 mol HCl
[HCl] = 0.01250 mol / 0.07570 L = 0.1651 M HCl
pH = -log (0.1651) = 0.78
