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Chapter 4: Problem 10

Itwenty-five \(\mathrm{mL}\) of a \(0.388 \mathrm{M}\) solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is mixed with \(35.3 \mathrm{~mL}\) of \(0.229 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\). What is the molarity of the resulting solution? Assume that the volumes are additive.

Short Answer

Expert verified
Answer: The molarity of the resulting solution is approximately 0.295 M.

Step by step solution

01

Calculate the moles of solute in each solution

To calculate the moles of Na2SO4 in each solution, we will multiply the volume of each solution by its molarity. For the first solution, 25mL of 0.388M Na2SO4: Moles of Na2SO4 = M1 × V1 Moles of Na2SO4 = 0.388 M × 0.025 L = 0.0097 moles For the second solution, 35.3mL of 0.229M Na2SO4: Moles of Na2SO4 = M2 × V2 Moles of Na2SO4 = 0.229 M × 0.0353 L = 0.00808 moles
02

Calculate the total moles of solute in the resulting solution

Add the moles of Na2SO4 found in step 1 from both solutions to find the total moles of Na2SO4 in the resulting solution: Total moles of Na2SO4 = moles from solution 1 + moles from solution 2 Total moles of Na2SO4 = 0.0097 moles + 0.00808 moles = 0.01778 moles
03

Calculate the total volume of the resulting solution

Since the volumes of the two solutions are additive, sum their volumes to find the total volume: Total volume = Volume of Solution 1 + Volume of Solution 2 Total volume = 0.025 L + 0.0353 L = 0.0603 L
04

Calculate the molarity of the resulting solution

Divide the total moles of the solute (Na2SO4) by the total volume of the solution to find the molarity of the resulting solution: Resulting molarity (M) = Total moles of Na2SO4 ÷ Total volume Resulting molarity (M) = 0.01778 moles ÷ 0.0603 L ≈ 0.295 M The molarity of the resulting solution is approximately 0.295 M.

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Most popular questions from this chapter

For an acid-base reaction, what is the reacting species (the ion or molecule that appears in the chemical equation) in the following bases? (a) barium hydroxide (b) trimethylamine \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\) (c) aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (d) sodium hydroxide

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