/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 114 You have \(1.00 \mathrm{~L}\) of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You have \(1.00 \mathrm{~L}\) of \(0.10-\mathrm{M}\) formic acid, \(\mathrm{HCOOH},\) whose \(K_{\mathrm{a}}=3.0 \times 10^{-4}\). You want to bubble into the formic acid solution sufficient HCl gas to decrease the pH of the formic acid solution by \(1.0 \mathrm{pH}\) unit. Calculate the volume of HCl (liters) that must be used at STP to bring about the desired change in pH. Assume no volume change has occurred in the solution due to the addition of HCl gas.

Short Answer

Expert verified
The volume of HCl gas needed is 3.43 L.

Step by step solution

01

Initial pH Calculation

First, calculate the initial pH of the formic acid solution. The expression for the dissociation of formic acid is \( \text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^- \). Using the initial concentration and the dissociation constant, set up the equilibrium expression: \( K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]} = 3.0 \times 10^{-4} \). Assume \([\text{H}^+] = x\). Then, \( x^2 = K_a \times (0.10 - x) \approx K_a \times 0.10 \). Solve for \( x \): \( x = \sqrt{3.0 \times 10^{-4} \times 0.10} = 1.7 \times 10^{-2} \). The initial pH is \( -\log(1.7 \times 10^{-2}) \approx 1.77 \).
02

Desired pH Calculation

Subtract 1.0 from the initial pH to find the desired pH. The desired pH is \( 1.77 - 1.0 = 0.77 \). Convert it back to hydrogen ion concentration: \([\text{H}^+] = 10^{-0.77} \approx 1.7 \times 10^{-1}\).
03

Change in Hydrogen Ion Concentration

Calculate the change in hydrogen ion concentration required. \( \Delta[\text{H}^+] = [\text{H}^+_{\text{final}}] - [\text{H}^+_{\text{initial}}] = 1.7 \times 10^{-1} - 1.7 \times 10^{-2} = 1.53 \times 10^{-1} \text{ M}\).
04

Calculate Moles of HCl Required

Determine the moles of HCl needed to achieve this change in hydrogen ion concentration. Since there is no change in volume, \( \Delta[\text{H}^+] = \text{moles HCl}/1.00 \). Thus, the moles of HCl needed is \( 1.53 \times 10^{-1} \text{ moles}\).
05

Convert Moles of HCl to Volume

Convert the moles of HCl to volume at STP. At STP, 1 mole of gas occupies 22.4 L. Thus, the volume of HCl is \( \frac{1.53 \times 10^{-1}}{1} \times 22.4 = 3.43 \text{ L} \).

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding pH Calculation
Calculating pH is fundamental in understanding acid-base equilibrium. pH is a measure of the acidity or basicity of a solution. It is defined as the negative logarithm of the hydrogen ion concentration: \( \text{pH} = -\log [\text{H}^+] \). This means that if you know the concentration of hydrogen ions in a solution, you can find its pH by using a calculator to compute the logarithm.
For instance, in this exercise, the concentration was calculated to be \(1.7 \times 10^{-2}\) M. By plugging this value into the formula, we get an initial pH of approximately 1.77. Understanding how to manipulate this formula is crucial when adjusting the pH of solutions, such as adding HCl to decrease pH.
  • Low pH values (< 7) indicate acidic solutions.
  • pH of exactly 7 signifies a neutral solution.
  • High pH values (> 7) indicate basic solutions.
When the pH decreases by 1 unit, the hydrogen ion concentration increases tenfold. This is critical to grasp when predicting how much acid is needed to achieve desired changes in pH.
Exploring Equilibrium Constant
The equilibrium constant, especially the acid dissociation constant \(K_a\), is key to understanding how acids ionize in solution. The \(K_a\) value measures the strength of an acid in solution. It tells us how well an acid can donate protons to a base. Formic acid has a \(K_a\) of \(3.0 \times 10^{-4}\), indicating it's a weak acid because it ionizes partially in water.
To find the hydrogen ion concentration, we use the formula: \[\text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^-\]\( K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]}\)
This equilibrium equation helps set up the relationship between the concentrations of reactants and products at equilibrium. By solving this equation, we find the concentration of \([\text{H}^+]\) at equilibrium. It’s important to apply simplifying assumptions, like \([\text{HCOOH}]\) remaining mostly unchanged, when the acid is weak and the equilibrium concentration is small compared to the initial concentration.
This concept is pivotal when predicting how a system responds to changes, such as adding more acid to shift the equilibrium.
All About Hydrogen Ion Concentration
Hydrogen ion concentration \([\text{H}^+]\) is a direct measure of a solution's acidity. The greater this concentration, the more acidic the solution. In the given exercise, changes in \([\text{H}^+]\) are essential to understand how much HCl gas is needed to achieve the desired pH.
Initially, the concentration was found to be \(1.7 \times 10^{-2}\) M. With a target pH drop of 1 unit, the \([\text{H}^+]\) needs to increase to approximately \(1.7 \times 10^{-1}\) M. The difference, \(1.53 \times 10^{-1}\) M, indicates how much more hydrogen ion concentration is needed.
This increase can be achieved through the addition of HCl, which dissociates completely in solution, directly contributing to the \([\text{H}^+]\). Hence, the volume of gas needed is calculated based on this change.
Ensuring clarity around these concentration changes is crucial for effective pH regulation, allowing calculations to directly inform how chemical additions will impact a system.

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

A buffer solution was prepared by adding \(4.95 \mathrm{~g}\) sodium acetate to \(250 . \mathrm{mL}\) of \(0.150-\mathrm{M}\) acetic acid. (a) What ions and molecules are present in the solution? List them in order of decreasing concentration. (b) Calculate the \(\mathrm{pH}\) of the buffer solution. (c) Calculate the pH of 100 . mL of the buffer solution if you add \(80 . \mathrm{mg} \mathrm{NaOH}\). (Assume negligible change in volume.) (d) Write a net ionic equation for the reaction that occurs to change the pH.

Write a balanced chemical equation for the equilibrium occurring when each of these solutes is added to water, then write the \(K_{\mathrm{sp}}\) expression. (a) \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) (b) Silver sulfate (c) Calcium phosphate (d) Manganese(III) hydroxide (e) Iron(II) carbonate

Calculate the \(\mathrm{pH}\) change when \(10.0 \mathrm{~mL}\) of \(0.100-\mathrm{M}\) \(\mathrm{NaOH}\) is added to \(90.0 \mathrm{~mL}\) pure water, and compare the \(\mathrm{pH}\) change with that when the same amount of \(\mathrm{NaOH}\) solution is added to \(90.0 \mathrm{~mL}\) of a buffer consisting of \(1.00-\mathrm{M} \mathrm{NH}_{3}\) and \(1.00-\mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\). Assume that the vol- umes are additive. \(K_{\mathrm{b}}\) of \(\mathrm{NH}_{3}=1.8 \times 10^{-5}\)

Write the chemical equation for the formation of each complex ion and write its formation constant expression. (a) \(\left[\mathrm{CoCl}_{6}\right]^{3-}\) (b) \(\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}\)

Calculate the \(K_{\mathrm{sp}}\) of \(\mathrm{HgI}_{2}\) given that its solubility in water is \(2.0 \times 10^{-10} \mathrm{M}\). Assume that there are no reactions other than the \(K_{\mathrm{sp}}\) reaction.

See all solutions

Recommended explanations on Chemistry Textbooks

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.