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What volume of 0.365 N lithium hydroxide is needed to completely neutralize 27.68 mL of 0.559 N nitric acid?

1 Answer

Jan 2, 2016

#LiOH(aq) + HNO_3(aq) rarr H_2O(l) + LiNO_3(aq)#

Explanation:

Moles of nitric acid?

#27.68xx10^(-3)cancelLxx0.559*mol*cancel(L^(-1))# #=# #1.54xx10^(-2)# #mol#.

So we need #(1.54xx10^(-2)*cancel(mol))/(0.365*cancel(mol)*L^-1)# #=# #4.24xx10^(-2)# #L#

Or #4.24xx10^(-2)# #cancelL# #xx# #1000# #mL# #cancel(L^-1)# #=# #??# #mL#

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