Question #7013e

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Question #7013e

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Answer 1 · 2017-04-22T22:20:52

Refer to the explanation

Explanation:

First, find the moles of $N a O H$ $NaOH$ using the formula of $n = c \cdot v$ $n = c*v$
number of moles = $\frac{v o l u m e}{1000} \cdot c o n c e n t r a t i o n$ $(volume)/1000 *concentration$

moles = $\frac{40}{1000} \cdot 0.2$ $(40)/1000*0.2$

moles = 0.008

Then here is the balanced equation of the neutralisation reaction while having the unknown dibasic acid.

$H_{2} X + 2 N a O H \to 2 H_{2} O + N a_{2} X$ $H_2X + 2NaOH -> 2H_2O + Na_2X$

Mole ratio is 1:2 so you divide the moles of the NaOH by 2.

$\frac{0.008}{2}$ $0.008/2$ =0.004

Then we use the formula moles = $\frac{m a s s}{A_{r} or M_{r}}$ $(mass)/(A_r or M_r)$

$M_{r} = \frac{m a s s}{m}$ $M_r = (mass)/ (m)$

$\frac{0.2}{0.004}$ $0.2/(0.004)$ = 50

Relative molecular mass of acid A is 50

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Question #7013e

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