Question

Question #c434b

Answer 1

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.