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Answer 1 · 2018-02-13T19:42:38

$sf(M_r=90)$

If your acid is monobasic then the equation is:

$sf(HA+NaOHrarrNaA+H_2O)$

$:.$ 1 mole $sf(HA-=)$ 1 mole $sf(NaOH)$

Normality is an obsolete concept so I will use molarity which, in this case, is the same.

$sf(c=n/v)$

$:.$ $sf(n=cxxv)$

$:.$ $sf(n_(NaOH)=1/(12)xx10/(1000)=8.333xx10^(-4))$

The number of moles of $sf(HA)$ must be the same so we can say:

$sf(n_(HA)=8.333xx10^(-4))$

$sf(m=nxxM_r)$

$:.$ $sf(M_r=m/n=0.075/(8.333xx10^(-4))$

$sf(M_r= 90)$