Question

A solution with a volume of 1.00 l is 0.450 m in ch 3 cooh(aq) and 0.550 m in ch 3 coona(aq). what will the ph be after 0.0800 mol of hcl is added to the solution?

Answer 1

Before addition of HCl,

conc. of CH3COOH = 0.450 M
conc. of CH3COONa = 0.550 M

After addition of 0.08 M HCl, following reaction occurs in system:
HCl + CH3COONa ↔ CH3COOH + NaCl

Thus, in reaction system conc. of CH3COOH will increase to 0.53 M (0.08M + 0.450M)
And, conc to CH3COONa will reduce to 0.47 M (0.550M - 0.08M)

Now, conc. of H+ ions = ka
`([acid])/([conjugated base])`
where ka = dissociation constant for acid = 10^-5 for Ch3COOH

∴ conc. of H+ ions =
`(0.53)/(0.47)`
= 1.1277 x 10^-5

Now, pH = -log [H+] = -log (1.1227 x 10^-5) = 4.94