Mole ConceptHard
Question
A quantity of 23.6 g of succinic acid is dissolved in 500 ml of 0.1 M acetic acid solution. Assuming that neither acid is dissociated in solution, calculate the molarity of ‘–COOH’ in the solution.
Options
A.0.3 M
B.0.5 M
C.0.9 M
D.0.8 M
Solution
Moles of succinic acid = $\frac{23.6}{118} = 0.2$
∴ Final molarity of –COOH group $= \frac{0.2 \times 2}{500} \times 1000 + 0.1 \times 1 = 0.9\text{ M}$
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