/*! 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 195 Match the following: Column I ... [FREE SOLUTION] | 91影视

91影视

Match the following: Column I (Buffer) \(\quad\) Column II A. \(1 \times 10^{-8} \mathrm{M} \mathrm{KOH}\) (p) \(1.3\) B. \(1 \times 10^{-8} \mathrm{M} \mathrm{HBr}\) (q) \(\mathrm{pH}=6.95\) C. \(1 \mathrm{M} \mathrm{HCl}+1 \mathrm{M}\) (r) \(\mathrm{pH}=7.0414\) \(\mathrm{NaOH}\) D. \(0.02 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) (s) \(\mathrm{pH}=7\) (t) acidic solution

Short Answer

Expert verified
A matches (s); B matches (t); C matches (s); D matches (t).

Step by step solution

01

Understand the Problem

We need to find the pH value corresponding to each solution in Column I and match each with the options in Column II.
02

Evaluate Option A - 1 脳 10鈦烩伕 M KOH

Since KOH is a strong base, it completely dissociates in solution. However, the concentration is very low. The presence of 10鈦烩伕 M KOH is comparable to the concentration of ext{OH}鈦 from water itself (10鈦烩伔 M). Thus, the pH = 7, which matches (s).
03

Evaluate Option B - 1 脳 10鈦烩伕 M HBr

HBr is a strong acid, so it will also dissociate completely. Like with KOH, the concentration is so low that the solution's pH is influenced by the neutrality of water, maintaining a pH approximately around 7, which is slightly acidic. Thus, match with (t).
04

Evaluate Option C - 1 M HCl + 1 M NaOH

This is a neutralization reaction resulting in water and salt. Since equal concentrations and volumes of a strong acid (HCl) and a strong base (NaOH) are mixed, they neutralize each other leading to a neutral solution with a pH of 7, matching (s).
05

Evaluate Option D - 0.02 M H鈧係O鈧

H鈧係O鈧 is a strong acid which dissociates completely in the first step to release H鈦 ions, resulting in a low pH value. 0.02 M H鈧係O鈧 will have a significant acidic character. This corresponds to a generic descriptor of an acidic solution, matching (t).
06

Match each option with Column II

- A matches with (s) because pH = 7 for 1 脳 10鈦烩伕 M KOH (very dilute solution makes it act neutral). - B matches with (t) since it's a very dilute acidic solution but leans slightly acidic. - C matches with (s), resulting in a neutral pH = 7 from equal amounts of HCl and NaOH. - D matches with (t) as it results in a more acidic solution.

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 91影视!

Key Concepts

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

pH calculation
The concept of pH is a cornerstone in understanding buffer solutions and acid-base chemistry. pH measures the concentration of hydrogen ions \(\text{H}^+\) in a solution, with a lower pH indicating higher acidity and a higher pH indicating higher alkalinity. The pH scale is logarithmic, ranging from 0 to 14. A pH of 7 is considered neutral, which is the pH of pure water. In the original problem, the solutions' pH was determined based on their ability to donate or accept protons. For very dilute solutions, like 1 脳 10鈦烩伕 M KOH, the pH nears that of neutral water, accounting for the natural ionization of water. This makes the pH closer to 7 due to the water's self-ionization, which contributes \(10^{-7}\) M \(\text{H}^+\) and \(\text{OH}^-\) each. This means such a dilution makes its contribution negligent against the natural ion balance of water.The real-life applicative value of understanding pH calculations is to predict the behavior of solutions in reactions and their interactions with biological systems. When working with pH, it's often practical to use the formula: \[\text{pH} = -\log [\text{H}^+]\] This formula helps calculate pH based on the concentration of hydrogen ions, making it easier to predict whether a solution is acidic, neutral, or basic.
neutralization reactions
Neutralization reactions are fantastic processes that occur when an acid and a base react to form water and a salt. This is of immense significance in acid-base chemistry because it perfectly illustrates how solutions balance each other out to attain neutrality. In the textbook solution, 1 M \(\text{HCl}\) and 1 M \(\text{NaOH}\) undergo a neutralization reaction. Because they are mixed in equal molar concentrations, they totally neutralize each other, resulting in a neutral pH of 7. This results in a solution that is neither acidic nor basic, consisting mainly of water and \(\text{NaCl}\), illustrating the principle of balancing acidity with basicity.During a neutralization reaction:
  • An acid donates protons (\(\text{H}^+\))
  • A base accepts protons
  • The result is water (\(\text{H}_2\text{O}\)) and a salt
These reactions are crucial in various industries, such as pharmaceuticals and environmental science, where maintaining a specific pH is necessary for the process or ecosystem stability. Understanding how these reactions work allows chemists and biologists to manipulate experimental and environmental conditions effectively.
acid-base chemistry
Acid-base chemistry is a fascinating field often central to understanding many other chemical processes. It involves the study of reactions between acids and bases and their roles in different chemical behaviors and equilibrium.In the given exercise, a series of solutions containing acids like \(\text{HBr}\) and \(\text{H}_2\text{SO}_4\) were examined. \(\text{HBr}\) and \(\text{H}_2\text{SO}_4\) are examples of strong acids that fully dissociate in solution, releasing \(\text{H}^+\) ions. Knowing how acids and bases interact helps in predicting the pH of a solution and its reactive properties.Furthermore, in the context of buffer solutions, acid-base chemistry allows the maintenance of a stable pH upon addition of small amounts of acid or base. Buffer solutions are crucial in biological systems as they prevent drastic changes in pH that can disrupt biological processes. It is this balance between the acidic and basic components of buffer solutions that prevents changes in pH levels, making acid-base reactions vital to life and industrial applications alike. By grasping acid-base chemistry fundamentals, students can better understand how substances interact in solutions and predict their behaviors as they form new compounds.

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

What is the correct sequence of active masses in increasing order in gaseous mixture containing one gram per litre of each of the following? I. \(\mathrm{NH}_{3}\) II. \(\mathrm{N}_{2}\) III. \(\mathrm{H}_{2}\) IV. \(\mathrm{O}_{2}\) Select the correct answer using the codes given below: a. III, I, IV, II b. III, IV, II, I c. II, I, IV, III d. IV, II, I, III

At constant temperature, the equilibrium constant (Kp) for the decomposition reaction \(\mathrm{N}_{2} \mathrm{O}_{4} \leftrightarrow 2 \mathrm{NO}_{2}\) is expressed by \(\mathrm{Kp}=\left(4 \mathrm{x}^{2} \mathrm{P}\right) /\left(1-\mathrm{x}^{2}\right)\), where \(\mathrm{P}=\) pressure \(x=\) extent of decomposition. Which one of the following statements is true? a. Kp increases with increase of \(\mathrm{P}\) b. Kp increases with increase of \(\mathrm{x}\) c. Kp increases with decrease of \(x\) d. Kp remains constant with change in P and \(x\).

Match the following: Column I Column II A. \(0.5 \mathrm{M} \mathrm{NO}_{3}+0.1 \mathrm{M} \quad\) (p) 7 \(\mathrm{NH}_{4} \mathrm{OH}\) B. \(0.1 \mathrm{M} \mathrm{KCl}+0.1 \mathrm{M} \quad\) (q) greater then 7 \(\mathrm{KNO}_{3}\) C. \(0.2 \mathrm{M} \mathrm{NaOH}+0.5 \mathrm{M}(\mathrm{r})\) between 1 to 7 \(\mathrm{HCOOH}\) D. \(0.1 \mathrm{NH}_{4} \mathrm{Cl}+0.1 \mathrm{M}(\mathrm{s}) 0.7\) \(\mathrm{KOH}\)

Match the following: Column I \(\quad\) Column II (p) Increase of pressure A. \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g})\) favours forward reaction B. \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow\) (q) Constant pressure \(2 \mathrm{NO}(\mathrm{g})\) favours forward reactior C. \(\mathrm{NH}_{2} \mathrm{COONH}_{4}(\mathrm{~s}) \rightleftharpoons\) (r) D. \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s}) \quad(\mathrm{s})\) \(+\mathrm{CO}_{2}(\mathrm{~g})\)

\((\mathrm{A}): \mathrm{H}_{3} \mathrm{PO}_{3}\) is a dibasic acid. \((\mathbf{R})\) : There are two hydrogen atoms directly attached to phosphorus

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.