Log out Go to App
Go to App

Learning Materials

Features

Discover

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The standard set by OSHA for the maximum amount of ammonia permitted in the workplace is \(5.00 \times 10^{-3} \%\) by mass. To determine a factory's compliance, \(10.00 \mathrm{~L}\) of air \((d=1.19 \mathrm{~g} / \mathrm{L})\) is bubbled into \(100.0 \mathrm{~mL}\) of \(0.02500 \mathrm{M} \mathrm{HCl}\) at the same temperature and pressure. Ammonia in the air bubbled in reacts with \(\mathrm{H}^{+}\) as follows: $$ \mathrm{NH}_{3}(a q)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{NH}_{4}^{+}(a q) $$ The unreacted hydrogen ions required \(57.00 \mathrm{~mL}\) of \(0.03500 \mathrm{M} \mathrm{NaOH}\) for complete neutralization. Is the factory compliant with the OSHA standards for ammonia in the workplace?

Write balanced equations for the following reactions in acid solution. (a) \(\mathrm{P}_{4}(s)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{PH}_{3}(g)+\mathrm{Cl}_{2}(g)\) (b) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{NO}_{2}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{NO}_{3}^{-}(a q)\) (c) \(\mathrm{HBrO}_{3}(a q)+\mathrm{Bi}(s) \longrightarrow \mathrm{HBrO}_{2}(a q)+\mathrm{Bi}_{2} \mathrm{O}_{3}(s)\) (d) \(\mathrm{CrO}_{4}^{2-}(a q)+\mathrm{SO}_{3}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\)

Household ammonia used for cleaning contains about \(10 \mathrm{~g}\) (two significant figures) of \(\mathrm{NH}_{3}\) in \(100 \mathrm{~mL}\) (two significant figures) of solution. What is the molarity of the \(\mathrm{NH}_{3}\) in solution?

Three students titrate different samples of the same solution of \(\mathrm{HCl}\) to obtain its molarity. Below are their data. Student \(\mathrm{A}: \quad 20.00 \mathrm{~mL} \mathrm{HCl}+20.00 \mathrm{~mL} \mathrm{H}_{2} \mathrm{O}\) \(0.100 \mathrm{M} \mathrm{NaOH}\) used to titrate to the equivalence point Student B: \(\quad 20.00 \mathrm{~mL} \mathrm{HCl}+40.00 \mathrm{~mL} \mathrm{H}_{2} \mathrm{O}\) \(0.100 \mathrm{M} \mathrm{NaOH}\) used to titrate to the equivalence point Student C: \(\quad 20.00 \mathrm{~mL} \mathrm{HCl}+20.00 \mathrm{~mL} \mathrm{H}_{2} \mathrm{O}\) \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) used to titrate to the equivalence point. All the students calculated the molarities correctly. Which (if any) of the following statements are true? (a) The molarity calculated by \(A\) is half that calculated by \(B\). (b) The molarity calculated by \(\mathrm{A}\) is equal to that calculated by \(\mathrm{C}\). (c) The molarity calculated by B is twice that calculated by C. (d) The molarity calculated by \(\mathrm{A}\) is twice that calculated by \(\mathrm{B}\). (e) The molarity calculated by \(\mathrm{A}\) is equal to that calculated by \(\mathrm{B}\). Challenge Problems

Laws passed in some states define a drunk driver as one who drives with a blood alcohol level of \(0.10 \%\) by mass or higher. The level of alcohol can be determined by titrating blood plasma with potassium dichromate according to the unbalanced equation $$ \mathrm{H}^{+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} $$ Assuming that the only substance that reacts with dichromate in blood plasma is alcohol, is a person legally drunk if \(38.94 \mathrm{~mL}\) of \(0.0723 \mathrm{M}\) potassium dichromate is required to titrate a \(50.0\) -g sample of blood plasma?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free