What is the ph of a solution (500 ml, 3.40 mm hoac) in which 50 ml of 1.00 m hcl is added? what percentage of the hoac is dissociated?

a) pH = 1.0414

b) ∝ = 0.156 %

Explanation:

∴ n HCL = (0.05L)*( 1 mol/L) = 0.05 mol

       HCl → H3O+  +  Cl-

ni     0.05     -              -

nf      -         0.05       0.05

weak acid:

∴ mol CH3COOH = (0.500L)*(3.40 E-3 mol/L ) = 1.7 E-3 mol

    CH3COOH ↔ CH3COO- +   H3O+

ni     1.7 E-3                 -                0.05

nf    1.7 E-3 - X            X             0.05 + X ≅ 0.05

∴ Ka = 1.75 E- 5 = ([ H3O+]*[CH3COO-])/[CH3COOH]

by lechatelier the reaction shifts to the left, then the moles of H3O+ in the weak acid is practically equal to 0.05 moles of the strong acid.

⇒  [H3O+] = 0.05 mol / 0.55 L = 0.0909 M

⇒ pH = - Log ( 0.0909 )

⇒ pH = 1.0414

replacing en Ka:

⇒ 1.75 E-5 = ((0.05+X)/0.55)*(X/0.55)) / ((1.7 E-3-X)/0.55)

⇒ 0.55X² + 0.3025X - 8.99 E-9 = 0

⇒ X = 4.8424 E-6 M

⇒ [H3O+] = 0.05mol/0.55L + 4.8424 E-6 = 0.0909 M

the assumption has been demonstrated

∴ percentage of tha CH3COOH is dissociated (∝):

⇒ ∝ = ( [CH3COO-] / ([CH3COO-] + [CH3COOH]) ) * 100

∴ [CH3COO-] = X = 4.824 E-6 M

∴ [CH3COOH] = 1.7 E-3 mol/0.55 L - 4.824 E-6 = 3.086 E-3 M

⇒ ∝ = ((4.824 E-6) / (3.086 E-3 + 4.824 E-6 )) * 100

⇒ ∝ = 0.156 %

c) should be the answer.