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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")#