Question

Question #9ef0b

Answer 1

M = .085 `V_1 xx M_1 = V_2 xx M_2`

Explanation:

The volume times the molarity of acid `{H^+}` equals the volume times the molarity of base `{OH^-}`

Li OH is a mono hydroxide base. so the molarity of LiOH equals the molarity of the base `{OH^-)`

`H_2CO_3` is a bi hydrogen acid. so the molarity of the acid `H^+`
equals `2 xx M` the molarity of `H_2CO_3`

`V_1 xx M (H+) = V_2 xx M (OH^-)`

`47 ml xx 2 M (H^+) = 37.5ml xx 0.215`

M = `37.5 xx 0.215 /(47.0 xx 2)`

M = .0858

Answer 2

`sf(0.085color(white)(l)"mol/l"`

Explanation:

Start with the equation:

`sf(H_2CO_(3(aq))+2LiOH_((aq))rarrLi_2CO_(3(aq))++H_2O_((l)))`

This tells us that 1 mole of `sf(H_2CO_3)` will be neutralised by 2 moles of `sf(LiOH)`.

To find the number of moles of `sf(LiOH)` we can say that:

`sf(c=n/v)`

`:.` `sf(n_(LiOH)=cxxv= 0.215xx37.5/1000=0.008)`

From the equation we can say that:

`sf(n_(H_2CO_3)=0.008/2=0.004)`

`:.` `sf(c_(H_2CO_3)=n_(H_2CO_3)/v=0.004/(47.0/1000)=0.085color(white)(l)"mol/l")`