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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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Askelto2 Askelto2
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11 years ago
A 5.65 -g sample of a weak acid with Ka = 1.3 x 10^- 4 was combined with 5.10 mL of 5.80 M NaOH and the resulting solution was diluted to 750 mL. The measured pH of the solution was 4.15.

what is the molar mass of this weak acid?
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wrote...
#1
11 years ago
moles NaOH = 0.0051 L x 5.80 M=0.0296
HA + OH- = A- + H2O
moles A- = 0.0296
[A-]= 0.0296 / 0.750 L=0.0394 M
pKa = 3.89

4.15 - 3.89 =0.26
10^ 0.26 =1.82 = [0.0394] / [HA]

[HA]= 0.0216 M
moles HA at equilibirum = 0.0216 x 0.750 L= 0.0162
initial moles HA = 0.0162 +0.0296 =0.0458

MM = 5.65 / 0.0458 =123.4 g/mol
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