Consider the titration of 20.00 ML of KOH 0.01 mol/L with HNO3 0.01 mol/L. Calculate the PH of the titration solution after the addition of the following titration volumes:

Question

A. 0.00 ml C. 20.00 ml B. 19.99 ml D. 25.00 ml

Answer 1

You can do 0.00 mL can't you? That's just 0.01 M KOH solution. That gives you the OH and you convert to pH.

The 20.00 mL addition SHOULD tell you that is the equivalence point which means you have pure NaCl at the point in water. What do you think the pH is when you know that neither Na%+ nor Cl^- are hydrolyzed.

For the 19.99 mL. I'll do millimols to keep the number of zeros down..
millimols KOH = 20.00 x 0.01 = 0.2000
millimols HNO3 = mL x M = 19.99 x 0.01 = 0.1999
Excess KOH = 0.2000-0.1999 = ?
M KOH excess = millimols/mL = ?
Convert KOH M to OH (or pOH) and then to pH.

Answer 2

thank you so much :)))