Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 150

A 6.50-g sample of a diprotic acid requires 137.5 mL of a 0.750 M NaOH solution for complete neutralization. Determine the molar mass of the acid.

Short Answer

Expert verified
The molar mass of the diprotic acid can be calculated by first determining the moles of NaOH (0.1031 mol) used in the reaction, then finding the moles of the diprotic acid (0.05155 mol) based on the stoichiometry of the balanced equation. Finally, dividing the mass of the diprotic acid (6.50 g) by the moles of the acid results in a molar mass of 126.0 g/mol for the diprotic acid.

Step by step solution

01

Write the balanced chemical equation for the reaction.

The general formula for the reaction between a diprotic acid (H2A) and sodium hydroxide (NaOH) is: H2A + 2NaOH -> 2H2O + Na2A (This equation represents the complete neutralization of the diprotic acid and the formation of water and the sodium salt of the acid.)
02

Find the moles of NaOH.

We are given the volume and the concentration of the NaOH solution. To find the moles of NaOH, we will use the formula: moles of NaOH = volume (in liters) × concentration moles of NaOH = (137.5 mL × (1 L / 1000 mL)) × 0.750 M moles of NaOH = 0.1031 mol
03

Determine the moles of the diprotic acid.

Using the stoichiometry from the balanced chemical equation, we can determine the moles of the diprotic acid (H2A): 1 moles of H2A react with 2 moles of NaOH So, for the moles of NaOH calculated in Step 2, we can find the moles of H2A by dividing the moles of NaOH by 2: moles of H2A = moles of NaOH / 2 moles of H2A = 0.1031 mol / 2 moles of H2A = 0.05155 mol
04

Calculate the molar mass of the diprotic acid.

Now that we have the moles of the diprotic acid, we can calculate its molar mass using the formula: molar mass = mass of the acid / moles of the acid We are given the mass of the diprotic acid as 6.50 g. So: molar mass = 6.50 g / 0.05155 mol molar mass = 126.0 g/mol The molar mass of the diprotic acid is 126.0 g/mol.

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

Write net ionic equations for the reaction, if any, that occurs when aqueous solutions of the following are mixed. a. chromium(III) chloride and sodium hydroxide b. silver nitrate and ammonium carbonate c. copper(II) sulfate and mercury(I) nitrate d. strontium nitrate and potassium iodide

Zinc and magnesium metal each react with hydrochloric acid according to the following equations: $$\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ $$\mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ A 10.00-g mixture of zinc and magnesium is reacted with the stoichiometric amount of hydrochloric acid. The reaction mixture is then reacted with 156 mL of 3.00 M silver nitrate to produce the maximum possible amount of silver chloride. a. Determine the percent magnesium by mass in the original mixture. b. If 78.0 mL of HCl was added, what was the concentration of the HCl?

Gold metal will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. It will dissolve, however, in aqua regia, a mixture of the two concentrated acids. The products of the reaction are the AuCl, \(^{-}\) ion and gaseous NO. Write a balanced equation for the dissolution of gold in aqua regia.

A mixture contains only \(\mathrm{NaCl}\) and \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) . A \(1.45-\mathrm{g}\) sample of the mixture is dissolved in water and an excess of NaOH is added, producing a precipitate of \(\mathrm{Al}(\mathrm{OH})_{3}\) . The precipitate is filtered, dried, and weighed. The mass of the precipitate is 0.107 g. What is the mass percent of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in the sample?

What mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is required to precipitate all of the silver ions from 75.0 mL of a 0.100-M solution of \(\mathrm{AgNO}_{3} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

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