If it takes 72 mL of 9 M KOH to neutralize 415 mL of sulfuric acid (H2SO4) solution, what is the concentration of the H2SO4
solution?
M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

H₂SO₄ + 2KOH —> K₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ (nₐ) = 1

Mole ratio of the base, KOH (n₆) = 2

Finally, we shall determine the concentration of H₂SO₄. This can be obtained as follow:

Volume of base, KOH (V₆) = 72 mL

Concentration of base, KOH (C₆) = 9 M

Volume acid, H₂SO₄ (Vₐ) = 415 mL

Mole ratio of the acid, H₂SO₄ (nₐ) = 1

Mole ratio of the base, KOH (n₆) = 2

CₐVₐ / C₆V₆ = nₐ/n₆

Cₐ × 415 / 9 × 72 = 1/2

Cₐ × 415 / 648 = 1/2

Cross multiply

Cₐ × 415 × 2 = 648

Cₐ × 830 = 648

Divide both side by 830

Cₐ = 648 / 830

Therefore, the concentration of H₂SO₄ is 0.78 M

Learn more:

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

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