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Answer 1 · 2016-08-23T18:58:11

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

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 · 2016-08-23T22:09:13

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

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:

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

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