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lesliealvarado1022
29.07.2019 •
Chemistry
A35.0-ml sample of 0.150 m acetic acid (ch3cooh) is titrated with 0.150 m naoh solution. calculate the ph after 17.5 ml of base have been added. the ka for the acetic acid equilibrium is 1.8 x 10-5. chegg
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Ответ:
moles of CH3COOH = 35ml * 0.15 m/1000 =0.00525 mol
moles of NaOH = volume of NaOH*no.of moles of NaOH
= 17.5 ml * 0.15/1000 = 0.002625
SO the reaction after add the NaOH:
CH3COOH(aq) +OH- (aq) ↔ CH3COO-(aq) +H2O(l)
initial 0.00525 0 0
change - 0.002625 +0.002625 +0.002625
equilibrium 0.002625 0.002625 0.002625
When the total volume = 35ml _ 17.5ml = 52.5ml = 0.0525L
∴[CH3COOH] = 0.002625/0.0525 = 0.05m
and [CH3COO-]= 0.002625/0.0525= 0.05 m
when PKa = -㏒Ka
= -㏒1.8x10^-5 = 4.74
by substitution in the following formula:
PH = Pka + ㏒[CH3COO-]/[CH3COOH]
= 4.74 + ㏒(0.05/0.05) = 4.74
∴PH = 4.74
Ответ: