Answer:
1.3 × 10⁻¹¹ M
Explanation:
We are going to do 4 successive dilutions. In each dilution, we will apply the dilution rule.
C₁.V₁=C₂.V₂
where,
C₁ and V₁ are concentration and volume of the initial state
C₂ and V₂ are concentration and volume of the final state
First dilution
C₁ = 3.1 × 10⁻⁵ M V₁ = 1.00 mL C₂ = ? V₂ = 40.00mL
[tex]C_{2}=\frac{C_{1}.V_{1}}{V_{2}} =\frac{3.1\times 10^{-5}M \times 1.00mL }{40.00mL} =7.8 \times 10^{-7}M[/tex]
Second dilution
C₁ = 7.8 × 10⁻⁷ M V₁ = 1.00 mL C₂ = ? V₂ = 40.00mL
[tex]C_{2}=\frac{C_{1}.V_{1}}{V_{2}} =\frac{7.8 \times 10^{-7}M \times 1.00mL }{40.00mL} =2.0 \times 10^{-8}M[/tex]
Third dilution
C₁ = 2.0 × 10⁻⁸ M V₁ = 1.00 mL C₂ = ? V₂ = 40.00mL
[tex]C_{2}=\frac{C_{1}.V_{1}}{V_{2}} =\frac{2.0 \times 10^{-8}M \times 1.00mL }{40.00mL} =5.0 \times 10^{-10}M[/tex]
Fourth dilution
C₁ = 5.0 × 10⁻¹⁰ M V₁ = 1.00 mL C₂ = ? V₂ = 40.00mL
[tex]C_{2}=\frac{C_{1}.V_{1}}{V_{2}} =\frac{5.0 \times 10^{-10}M \times 1.00mL }{40.00mL} =1.3 \times 10^{-11}M[/tex]
