It was determined that a 0.10 M solution of an acid was only 2.5% ionkzed. Find the Ka and pKa for the acid.

HA + H2O---> H3O+ + A-

Ka=[H3O+][A-]/[HA]

..........HA I H3O+ I A-

I.........I 0.1 M I 0 I 0 I
C........I-2(0.1*0.025) I (0.1*0.025)I (0.1*0.025) I
E.........I 0.1-2(0.1*0.025) I (0.1*0.025) I (0.1*0.025)

Ka=[(0.1*0.025][(0.1*0.025]/[0.1-2(0.1*0.025]

pka=-log(ka)

Opps,

HA + H2O---> H3O+ + A-

Ka=[H3O+][A-]/[HA]

..........HA I H3O+ I A-

I.........I 0.1 M I 0 I 0 I
C........I-2(0.1*0.025) I (0.1*0.025)I (0.1*0.025) I
E.........I 0.1-(0.1*0.025) I (0.1*0.025) I (0.1*0.025)

Ka=[(0.1*0.025][(0.1*0.025]/[0.1-(0.1*0.025]

pka=-log(ka)

I apologize about that one.

To find the Ka (acid dissociation constant) and pKa (negative logarithm of Ka) for the acid, we can use several steps.

Step 1: Determine the concentration of the ionized form of the acid.
Since the 0.10 M solution is only 2.5% ionized, we can calculate the concentration of the ionized form using the equation:
Ionized concentration = 0.025 * 0.10 M = 0.0025 M

Step 2: Determine the concentration of the non-ionized form of the acid.
Since the acid is only 2.5% ionized, it means that 97.5% of the acid remains in the non-ionized form. Therefore, the concentration of the non-ionized form can be calculated as:
Non-ionized concentration = (1 - 0.025) * 0.10 M = 0.0975 M

Step 3: Write the equilibrium expression for the ionization of the acid.
The ionization of the acid can be represented as:
HA ⇌ H+ + A-
The equilibrium expression for this reaction is:
Ka = [H+][A-] / [HA]

Step 4: Substitute the values into the equilibrium expression and solve for Ka.
Using the concentrations we calculated in Step 1 and Step 2, we can plug them into the equation:
Ka = (0.0025 M)(0.0025 M) / (0.0975 M)
Ka = 6.41 x 10^-5

Step 5: Calculate pKa.
The pKa is the negative logarithm (base 10) of the Ka value:
pKa = -log(6.41 x 10^-5)
pKa = 4.19

Therefore, the Ka for the acid is 6.41 x 10^-5 and the pKa is 4.19.