Question #c434b

1 Answer
Sep 27, 2016

Answer:

Approx #0.2*mol*L^-1# with respect to #H_2SO_4#.

Given the stoichiometric equation, this is a #2C_2V_2=C_1V_1# equation.

Explanation:

As always we need a stoichiometrically balanced equation to show the molar equivalence.

#H_2SO_4(aq) + 2LiOH(aq) rarr Li_2SO_4(aq) + 2H_2O(l)#

As you already know, 2 equiv of lithium hydroxide react with 1 equiv sulfuric acid.

#"Moles of lithium hydroxide"# #=#

#83.6*mLxx10^-3*L*mL^-1xx0.12*mol*L^-1#

#=# #0.0100*mol#

And thus clearly, there were #(0.0100*mol)/2# #H_2SO_4# in the initial #25*mL# volume.

#"Concentration"# #=# #"Moles of sulfuric acid"/"Volume of solution"#

#=# #(5.02xx10^-3*mol)/(25xx10^-3L)# #=# #0.201*mol*L^-1# with respect to #H_2SO_4#.

Given the stoichiometric equation, this is a #2C_2V_2=C_1V_1# equation.