Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 103

What is the \(\mathrm{pH}\) of a \(0.200 \mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) ? You may assume that the first ionization is complete. The second ionization constant is \(0.010\).

Short Answer

Expert verified
Answer: The pH of the 0.200 M H2SO4 solution is approximately 0.68.

Step by step solution

01

Determine the concentration of H+ from the first ionization of H2SO4

Since we're given that the first ionization is complete, we can directly calculate the concentration of H+ ions produced in this step. For every one mole of H2SO4, one mole of H+ ions and one mole of HSO4- ions are produced. Therefore, the concentration of H+ ions from the first ionization is equal to the concentration of H2SO4: 0.200 M.
02

Determine the concentration of H+ from the second ionization of HSO4-

Now, we need to determine the concentration of H+ ions produced from the second ionization of HSO4-. The second ionization constant (K2) is given as 0.010. We can represent the second ionization as: HSO4- (aq) <=> H+ (aq) + SO42- (aq) Using the assumption that the initial concentration of H+ ions is entirely from the first ionization (as calculated in Step 1), we can set up the reaction quotient (Q) for the second ionization: Q = [H+][SO42-] / [HSO4-] Substituting K2 for Q and the initial concentrations into the equation, we get: 0.010 = ([0.200 + x] * x) / (0.200 - x) Where x represents the increase in [H+] and the decrease in [HSO4-] (i.e., the amount of HSO4- ionized in the second step).
03

Solve for x

To solve for x, we can first assume that x is small compared to 0.200 (since K2 is small). This simplifies our equation to: 0.010 = (0.200 * x) / 0.200 Solving for x, we get: x = 0.010 M This value represents the additional concentration of H+ ions formed from the second ionization of HSO4-.
04

Calculate the total concentration of H+ ions

Now, we can find the total concentration of H+ ions in the solution by adding the concentrations from both ionizations: [H+] = [H+ from first ionization] + [H+ from second ionization] [H+] = 0.200 M + 0.010 M [H+] = 0.210 M
05

Calculate the pH of the solution

Finally, we can calculate the pH of the solution using the formula: pH = -log10([H+]) pH = -log10(0.210) pH ≈ 0.68 So the pH of the 0.200 M H2SO4 solution is approximately 0.68.

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

Consider a weak base, \(\mathrm{NaB}(\mathrm{MM}=281 \mathrm{~g} / \mathrm{mol})\). An aqueous solution of NaB has a \(\mathrm{pH}\) of \(8.73\) and an osmotic pressure of \(55 \mathrm{~mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\). What is \(K_{b}\) for the weak base NaB?

Consider the process $$\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \quad \Delta H^{\circ}=55.8 \mathrm{~kJ}$$ (a) Will the \(\mathrm{pH}\) of pure water at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) be \(7.0 ?\) (b) If not, calculate the \(\mathrm{pH}\) of pure water at \(37^{\circ} \mathrm{C}\).

Using the Tables in Appendix 1, calculate \(\Delta H\) for the reaction of the following. (a) \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) with \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) (b) \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) with \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M}\) HF, taking the heat of formation of \(\mathrm{HF}(a q)\) to be \(-320.1 \mathrm{~kJ} / \mathrm{mol}\)

Write a balanced equation showing how the \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ion can be either a Bronsted-Lowry acid or a Bronsted-Lowry base.

Find \(\left[\mathrm{H}^{+}\right]\) and the \(\mathrm{pH}\) of the following solutions. (a) \(1.75 \mathrm{~L}\) of a \(37.5 \%\) (by mass) solution \((d=1.00 \mathrm{~g} / \mathrm{mL})\) of \(\mathrm{HCl} .\) What is the \(\mathrm{pH}\) of \(0.175 \mathrm{~L}\) of the same solution? (b) A solution made up of \(22 \mathrm{~g}\) of \(\mathrm{HBr}\) dissolved in enough water to make \(479 \mathrm{~mL}\) of solution. What is the \(\mathrm{pH}\) if the same mass of \(\mathrm{HBr}\) is dissolved in enough water to make \(47.9 \mathrm{~mL}\) of solution?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

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