HA -----> H(+) + A(-)
0,2-x.........x..........x
100% - 99,4% = 0,6% = 0,006
x = 0,6%×0,2 = 0,006 × 0,2 = 0,0012
HA --------------> H(+) + A(-)
0,2-x.....................x..........x
0,2-0,0012.....0,0012..0,0012
0,1988.............0,0012..0,0012
K = [H+][A-]/[HA]
K = (0,0012)^2/0,1988
K = 0,0000072434
pKa = -logK= -log0,0000072434 = 5,14
:)
Answer:
1.33 M
Explanation:
We'll begin by writing out the data obtained from the question. This includes the following:
Volume of the stock solution (V1) = 0.5L
Molarity of the stock solution (M1) = 4M
Volume of diluted solution (V2) = 1.5L
Molarity of the diluted solution (M2) =.?
With the application of the dilution formula, the molarity of the diluted solution can be obtained as follow:
M1V1 = M2V2
4 x 0.5 = M2 x 1.5
Divide both side by 1.5
M2 = (4 x 0.5) / 1.5
M2 = 1.33 M
Therefore the molarity of the diluted solution is 1.33 M
