Question

Question #b8da5

Answer 1

`sf(M_r=90)`

Explanation:

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)`