The ionization constant for dichloroacetic acid hc2ho2cl2 is 5.0 x 10^-2 . what is the ph of a 0.15 molar solution of this acid?

pH = 1.19

Explanation:

The formula for Ka is:   Ka = [H+][A-]/[HA]

where:  [H+] = concentration of H+ ions

[A-] = concentration of conjugate base ions

[HA] = concentration of undissociated acid molecules

Equation of reaction:  Cl₂CHCOOH ---> H+ + Cl₂CHCOO-

From the equation above, dichloroacetic acid dissociates one H+ ion for every Cl₂CHCOO- ion,

so [H+] = [Cl₂CHCOO-].

Let x represent the concentration of H+ that dissociates from HA, then [HA] = C - x where C is the initial concentration.

Substituting these values into the Ka equation:

Ka = x · x / (C -x)

Ka = x²/(C - x)

(C - x)Ka = x²

x² = CKa - Kax

x² + Kax - CKa = 0

Solve for x using the quadratic formula:

x = [-b ± √(b² - 4ac)]/2a

Note: There are two solutions for x. However only the positive value of x is used since x represents a concentration of ions in solution, and so cannot be negative.

x = [-Ka + √(Ka² + 4CKa)]/2

Substitute the values for Ka and C in the equation above:  

Ka = 5.0 x 10^-2

C = 0.15 M  

x = {-5.0 x 10^-2 + √[(5.0. x 10^-2)² + 4(0.15)(5.0 x 10^-2)]}/2

x = (-5.0 x 10^-2 + 1.80 x 10^-1)/2

x = 0.13/2

x = 6.50 x 10^-2

To find pH, we use the formula;

pH = -log[H+]

pH = -log(x)

pH = -log(6.50 x 10^-2)

pH = -(-1.19)

pH = 1.19

answer; the numbers of valence electrons in the acetic acid molecule (ch3co2h) is 24;