Ionic EquilibriumHard
Question
A 0.28 g sample of an unknown monoprotic organic acid is dissolved in water and titrated with a 0.1 M sodium hydroxide solution. After the addition of 17.5 ml of base, a pH of 5.0 is recorded. The equivalence point is reached when a total of 35.0 ml of NaOH is added. The molar mass of the organic acid is
Options
A.160
B.80
C.40
D.120
Solution
$n_{HA} = n_{OH^{-}} \Rightarrow \frac{0.28}{M} = \frac{35 \times 0.1}{1000} \Rightarrow M = 80$
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