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Answer 1 · 2017-12-03T11:41:18

(a) 38.73 ml

(b) 30.79 ml

Adipic acid is di - protic:

$sf(H_2X+2NaOHrarrNa_2X+2H_2O)$

$sf(M_r=146.14)$

The no. moles of $sf(H_2X)$ is given by:

$sf(n_(H_2X)=m/M_r=0.283/146.14=0.0019365)$

From the equation we can say that the no. moles of $sf(NaOH)$ must be twice that:

$sf(n_(NaOH)=2xx0.0019365=0.003873)$

$sf(c=n/v)$

$:.$ $sf(v=n/c=0.003873/0.1000=0.03373color(white)(x)L)$

$sf(v=38.73color(white)(x)ml)$

(b)

Now the acid is 74% pure:

$sf(m_(H_2X)=0.304xx0.74=0.22496color(white)(x)g)$

$sf(n_(H_2X)=m/M_r=0.22496/146.14=0.00153934)$

$:.$ $sf(n_(NaOH)=0.00153934xx2=0.00307869)$

$sf(v=n/c=0.0030786/0.1000=0.0307869color(white)(x)L)$

$sf(v=30.79color(white)(x)ml)$