# Question #e41ef

Dec 3, 2017

(a) 38.73 ml

(b) 30.79 ml

#### Explanation:

Adipic acid is di - protic:

$\textsf{{H}_{2} X + 2 N a O H \rightarrow N {a}_{2} X + 2 {H}_{2} O}$

$\textsf{{M}_{r} = 146.14}$

The no. moles of $\textsf{{H}_{2} X}$ is given by:

$\textsf{{n}_{{H}_{2} X} = \frac{m}{M} _ r = \frac{0.283}{146.14} = 0.0019365}$

From the equation we can say that the no. moles of $\textsf{N a O H}$ must be twice that:

$\textsf{{n}_{N a O H} = 2 \times 0.0019365 = 0.003873}$

$\textsf{c = \frac{n}{v}}$

$\therefore$$\textsf{v = \frac{n}{c} = \frac{0.003873}{0.1000} = 0.03373 \textcolor{w h i t e}{x} L}$

$\textsf{v = 38.73 \textcolor{w h i t e}{x} m l}$

(b)

Now the acid is 74% pure:

$\textsf{{m}_{{H}_{2} X} = 0.304 \times 0.74 = 0.22496 \textcolor{w h i t e}{x} g}$

$\textsf{{n}_{{H}_{2} X} = \frac{m}{M} _ r = \frac{0.22496}{146.14} = 0.00153934}$

$\therefore$$\textsf{{n}_{N a O H} = 0.00153934 \times 2 = 0.00307869}$

$\textsf{v = \frac{n}{c} = \frac{0.0030786}{0.1000} = 0.0307869 \textcolor{w h i t e}{x} L}$

$\textsf{v = 30.79 \textcolor{w h i t e}{x} m l}$