A 500.0 ml buffer solution contains 0.30 m acetic acid (ch3cooh) and 0.20 m sodium acetate(nach3coo). what will the ph of th is solution be after the addition of 20.0 ml of 1.00 m naohsolution? [ka(ch3cooh) = 1.8 × 10–5]
1 answer:
First, we need to get no.of moles of acetic acid = molarity * volume = 0.3 m * 0.5 L = 0.15 moles then, we need to get no. of moles acetate = molarity *volume = 0.2m * 0.5L = 0.1 moles and we have also to get moles of NaOH = molarity * volume = 1 m* 0.02L = 0.02 moles according to the reaction equation and by using ICE table: CH3COOH + OH- → CH3COO- + H2O ∴ moles acetic acid = 0.15 - 0.02 = 0.13 moles when [acetic acid] = moles / total volume = 0.13 moles / (0.5L + 0.02L) = 0.25 M and moles of acetate = 0.1 + 0.02 = 0.12 moles ∴[acetate] = moles / total volume = 0.12 moles / 0.52L = 0.23 M and when we have the value of Ka so we can get the Pka from this formula:Pka = -㏒ Ka = -㏒ 1.8 x 10^-5 = 4.7 by using H-H equation we can get the PH:∵PH = Pka + ㏒[acetate] /[acetic acid] ∴PH = 4.7 + ㏒(0.23 /0.25] = 4.66
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