HNO3 + KOH ==> KNO3 + H2O
millimols HNO3 = M x mL = approx 200 but this is just an estimate.
mmols KOH - 0.57 x 400 = approx 228.
So HNO3 is the limiting reagent; all of the HNO3 will be used and 28 mmols KOH will be left over. So the pH will be determined by the excess KOH.
M KOH = mmols/mL = approx 28/412.5 = ?, then pH + pOH = pKw = 14 and solve for pH.
Remember to recompute all of the numbers since I've estimated them.
How do i find the PH and pOH of 12.5ml of 15.8M HNO3 added to 400ml of 0.57M KOH
1 answer