How many moles of sodium acetate (nach3coo) must be added to 1.000 liter of a 0.500 m solution of acetic acid (ch3cooh) to produce a ph of 5.675? the ionization constant of acetic acid is 1.8 × 10−5 ?

ck(CH₃COOH) = 0,5 M.

V = 1 L.

Ka = 1,8·10⁻⁵.

pKa = -logKa = 4,75.

pH = 5,675.

Henderson–Hasselbalch equation: pH = pKa + log(cs/ck).
5,675 = 4,75 + log(cs/ck).
log(cs/ck) = 5,675 - 4,75 = 0,925.
cs/ck = 10∧(0,925) = 8,4.

cs = 8,4 · 0,5 M = 4,2 M.

n(CH₃COONa) = 4,2 mol/L · 1 L = 4,2 mol.

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