Answer:
a) pH = 1.0414
b) ∝ = 0.156 %
Step-by-step 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 %