Answer to Question #94673 in General Chemistry for Kyla

Question

What volume of 12.0 M sulfuric acid would be required to prepare 250 mL of 6.5 M sulfuric acid?

Answer 1

c₁(H₂SO₄) = 12.0M

c₂(H₂SO₄) = 6.5M

V₂(H₂SO₄) = 250 mL = 0.25 L

V₁(H₂SO₄) - ?

The amount of sulfuric acid (in moles) in the required volume of stock solution is the same as in the diluted sample:

n₁(H₂SO₄) = n₂(H₂SO₄)

c₁(H₂SO₄) V₁(H₂SO₄) = c₂(H₂SO₄) V₂(H₂SO₄)

$V_1(H_2SO_4)=\frac{c_2(H_2SO_4) \cdot V_2(H_2SO_4)}{c_1(H_2SO_4)}=\frac{6.5 \frac{mol}{L} \cdot 0.25 L}{12.0 \frac{mol}{L}}=0.135 L = 135 mL$