If 25.0 mL of 0.199 mol/L aqueous acid solution was extracted with 25.0 mL of octan-1-ol and the concentration of the aqueous acid was 0.143 mol/L after extraction, what was the concentration of the acid in the octan-1-ol after extraction?Question 6Answer0.3916 mol/L0.0560 mol/L0.0616 mol/L0.0672 mol/L
Question
If 25.0 mL of 0.199 mol/L aqueous acid solution was extracted with 25.0 mL of octan-1-ol and the concentration of the aqueous acid was 0.143 mol/L after extraction, what was the concentration of the acid in the octan-1-ol after extraction?Question 6Answer0.3916 mol/L0.0560 mol/L0.0616 mol/L0.0672 mol/L
Solution
To solve this problem, we first need to calculate the number of moles of acid before and after the extraction.
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Calculate the number of moles of acid before extraction: The number of moles = volume (L) x concentration (mol/L) The number of moles before extraction = 0.025 L x 0.199 mol/L = 0.004975 mol
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Calculate the number of moles of acid after extraction: The number of moles after extraction = 0.025 L x 0.143 mol/L = 0.003575 mol
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Calculate the number of moles of acid extracted: The number of moles extracted = moles before extraction - moles after extraction The number of moles extracted = 0.004975 mol - 0.003575 mol = 0.0014 mol
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Calculate the concentration of the acid in the octan-1-ol: The concentration = number of moles / volume (L) The concentration of the acid in the octan-1-ol = 0.0014 mol / 0.025 L = 0.056 mol/L
So, the concentration of the acid in the octan-1-ol after extraction was 0.056 mol/L.
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