# The titration of 20.0 mL of an unknown concentration H_2SO_4 solution requires 83.6 mL of 0.12 M LiOH solution. What is the concentration of the H_2SO_4 solution (in M)?

Jun 10, 2016

${\left({C}_{M}\right)}_{{H}_{2} S {O}_{4}} = 0.25 \setminus m o l . {L}^{-} 1$

#### Explanation:

Write a balanced chemical equation for the neutralization reaction to figure out the mole relationship between the sulfuric acid and lithium hydroxide.

${H}_{2} S {O}_{4} + 2 L i O H \to L {i}_{2} S {O}_{4} + 2 {H}_{2} O$

${n}_{{H}_{2} S {O}_{4}} = \frac{1}{2} \setminus {n}_{L i O H}$

${\underbrace{\left({C}_{M} \times V\right)}}_{\text{number of moles of "H_2SO_4) = 1/2\ \ \ \ \ \ xxunderbrace((C_MxxV))_("number of moles of } L i O H}$

Note that ${C}_{M}$ is the molar concentration and $V$is the volume of solution.

${\left({C}_{M} \times V\right)}_{{H}_{2} S {O}_{4}} = \frac{1}{2} {\left({C}_{M} \times V\right)}_{L i O H}$

(C_M)_(H_2SO_4) = (C_MxxV)_(LiOH)/( 2* V_(H_2SO_4)

${\left({C}_{M}\right)}_{{H}_{2} S {O}_{4}} = \frac{0.12 \setminus m o l . {L}^{-} 1 \times 83.6 \setminus m L}{2 \times 20.0 \setminus m L}$

${\left({C}_{M}\right)}_{{H}_{2} S {O}_{4}} = \frac{0.12 \setminus m o l . {L}^{-} 1 \times 83.6 \setminus \cancel{m L}}{2 \times 20.0 \setminus \cancel{m L}}$

${\left({C}_{M}\right)}_{{H}_{2} S {O}_{4}} = 0.25 \setminus m o l . {L}^{-} 1$