\"The HCl solution was made by mixing 20 mL concentrated HCl (12 M) with 100 mL

Question

"The HCl solution was made by mixing 20 mL concentrated HCl (12 M) with 100 mL isopropyl alcohol. Assuming there is no contraction or expansion of volume on mixing, what is the molarity of the resulting solution? How many mL should you need if all your starting material (5.9 mmol) has been converted?"

The molar ratio is 1:1:1, but I'm not sure what molecular weight or density I should use in a mixed solution like this. In this experiment, I also used 10 mL of 50/50 t-butylamine and N-methylpyrrolidinone which should have equated to 5.9mmol, but again am not sure what MW or density to use.

Thank you for any help!!

Explanation / Answer

Dilution of HCl stock

molarity of stock (M1) = 12 M

molarity of final solution (M2) = ? M

volume of stock taken (V1) = 20 ml

final volume of solution (V2)= 100 ml

thus,

M2 = M1V1/V2

molarity of final diluted HCl solution (M2) = 12 M x 20 ml/100 ml = 2.4 M

-

For a 5.9 mmol base we would need = 5.9 mmol/2.4 M = 2.46 ml HCl solution for complete neutralization

-

For a 10 ml (5 ml t-butylamine + 5 ml N-methylpyrrolidine) solution

So use density of t-butylamine for 5 ml and density of N-methylpyrroldiine for another 5 ml soluion.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.