Answer:
pH = 4.349
Explanation:
When HI is added to sodium acetate, CH3COONa, acetic acid is produced as follows:
CH3COONa + HI → CH3COOH + NaI
That means the moles of HI added are the moles of acetic acid produced and the moles of sodium acetate consumed.
Moles of HI:
0.0605L * (0.223mol/L) = 0.01349mol
To find pH of the buffer (Mixture of weak acid, acetic acid, and conjugate base, sodium acetate) we need to solve H-H euation:
pH = pKa + log [CH3COONa] / [CH3COOH]
Where pKa is -log of Ka of the acid: -log1.8x10⁻⁵ = 4.745
And [] could be taken as moles of each specie
Moles CH3COONa:
Inital moles: 0.086L * (0.385mol/L) = 0.03311
Final moles: 0.03311mol - 0.01349mol = 0.01962mol CH3COONa
Moles CH3COOH:
Inital moles: 0.086L * (0.410mol/L) = 0.03256
Final moles: 0.03256mol + 0.01349mol = 0.04875mol CH3COOH
pH = 4.745+ log [0.01962mol CH3COONa] / [0.04875mol CH3COOH]
pH = 4.3497
Rigth answer is:
