Question

CUAL SERIA EL PH DEL SISTEMA AMORTIGUADOR(CH3COOH 0.3 M/CH3COONa 0.3 M ) SI SE AGREGAN 5 ML DE NaOH 0.1 MKA=1.76X10^-5

Answer 1

CH3COONa ---------------- Na+ + CH3COO-

*CH3COOH + H20 ----------------- CH3COO- + H3O+

NaOH -------------- Na+ + OH-


1005 mL sol ___________ 0,3 mol CH3COONa
1000 mL sol ___________ X = 0,2985 mol CH3COONa

1005 mL sol ___________ 0,3 mol CH3COOH
1000 mL sol ___________ X = 0,2985 mol CH3COOH

1000 mL sol __________ 0,1 mol NaOH
5 mL sol _____________ X = 5x10-4 mol NaOH

1005 mL sol ___________  5x10-4 mol NaOH
1000 mL sol ___________ X = 4,98x10-4 mol NaOH


Ka = ([CH3COO-] + [H3O+]) . [H3O+] / ([CH3COOH] - [H3O+]) 

[H3O+] = Ka . ([CH3COOH] - [H3O+]) / ([CH3COO-] + [H3O+])

[H3O+] = 1,7x10-5 . ([0,2985] - [4,98x10-4]) / ([0,2985] + [4,98x10-4]

[H3O+] = 1,7x10-5 . 0.298 / 0,299 = 1,69x10-5

pH = -log [1,69x10-5] = 4,77

*Es un acido debil, por lo tanto, lleva una doble flecha.