You're adding HCl to the buffer solution of part 1.
question 1 gives NaAc = 0.10M
and HAc = 0.20M
You are taking 100 mL; theefore,
millimols NaAc = base = 0.1 x 100 = 10.
mmols HAc = acid = 0.20 x 100 = 20.
You are adding
28.8 mL x 0.035 = about 1 millimols HCl.
.........Ac^- + H^+ ==>HAc
I........10.....0.......20
add.............1..................
C........-1....-1.......+1
E.........9.....0.......21
1)A buffer solution that is 0.10 M sodium acetate and 0.20 M acetic acid is prepared. Calculate the initial pH of this solution.
The Ka for CH3COOH is 1.8 x 10-5 M. As usual, report pH to 2 decimal places.
*A buffered solution resists a change in pH.*
2)Calculate the pH when 28.8 mL of 0.035 M HCl is added to 100.0 mL of the above buffer.
I know I have to create an ICE table, but how do i set it up for question number two?
1 answer