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)?

Question

Avery Adams · Accepted Answer

#(C_M)_(H_2SO_4) = 0.25 \ mol.L^-1#

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

#H_2SO_4 + 2LiOH -> Li_2SO_4 + 2H_2O#

#n_(H_2SO_4) = 1/2 \ n_(LiOH)#

#underbrace((C_MxxV))_("number of moles of "H_2SO_4) = 1/2\ \ \ \ \ \ xxunderbrace((C_MxxV))_("number of moles of "LiOH)#

Note that #C_M# is the molar concentration and #V#is the volume of solution.

#(C_MxxV)_(H_2SO_4) = 1/2( C_MxxV)_(LiOH)#

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

#(C_M)_(H_2SO_4) = (0.12 \ mol.L^-1xx 83.6 \ mL)/( 2xx20.0 \ mL)#

#(C_M)_(H_2SO_4) = (0.12 \ mol.L^-1xx 83.6 \ cancel (mL))/( 2xx20.0 \ cancel(mL))#

#(C_M)_(H_2SO_4) = 0.25 \ mol.L^-1#

Jose Ayala · Answer

undefined To find the concentration of the H₂SO₄ solution, use the formula:

$$ M_1V_1 = M_2V_2 $$

Where:
- $ M_1 $ is the concentration of the H₂SO₄ solution (in M)
- $ V_1 $ is the volume of the H₂SO₄ solution used in the titration (in mL)
- $ M_2 $ is the concentration of the LiOH solution (in M)
- $ V_2 $ is the volume of the LiOH solution used in the titration (in mL)

Given:
- $ V_1 = 20.0 $ mL
- $ M_2 = 0.12 $ M
- $ V_2 = 83.6 $ mL

Substitute the values into the formula and solve for $ M_1 $:

$$ M_1 = \frac{{M_2V_2}}{{V_1}} $$

$$ M_1 = \frac{{0.12 \, \text{M} \times 83.6 \, \text{mL}}}{{20.0 \, \text{mL}}} $$

$$ M_1 = \frac{{10.032}}{{20.0}} \, \text{M} $$

$$ M_1 = 0.502 \, \text{M} $$

Therefore, the concentration of the H₂SO₄ solution is $ 0.502 \, \text{M} $.