Acid and Base?

Question

The initial solution in a titration is 20ml of 0.2M HNO3. The nitric acid is titrated with 0.15M KOH in 5.0ml volumes.

1) What is the initial pH of the nitric acid?
2) What is the pH after the addition of 10.0ml KOH
3) How many times of titration are required to reach equivalence point

Answer 1

1) HNO3-->H+ NO3-
strong electrolyte completely dissociated
[H+] =0.2
pH = log1/[H+]= log 1/0.2=log 5 =0.7

2) HNO3+KOH---> KNO3+H2O
moles in 20ml of 0.2M = 20*(10^-3)*0.2 = 4*10^-3mole
10mL of 0.15M KOH = 0.15*0.01=0.0015=1.5 *10^-3mole
So, you obtain1.510^-3 mole of KNO3
and have( 4-1.5)10^-3 mole of HNO3 left (this in a volume of10+20 =30ml
Concentraton of HNO3 unreacted= 2.5*10^-3/0.03=0.083M
pH = log 1/0.083=log12=1.08

3)to neutralize, you need 20*0.2/0.15 =26.6mL
you need 6 times at least to reach equivalence point

Answer 2

1. Since HNO3 is a strong acid, [H+] = 0.2 M. The pH is -log 0.2.

2. You're starting with 0.004 moles of HNO3, and you've added 0.003 moles of KOH. That leave 0.001 mole HNO3 in a total volume of 0.03 L. Calculate the concentration of HNO3 and then calculate the pH from that.

3. Figure out how many 5 mL additions of KOH will contain 0.004 moles of KOH.