## Data on Weak Acids and Weak Bases### Acid| **Name** | **Formula** | **Ka** ||----------------|--------------|-------------------|| Nitrous Acid | HNO₂ | 4.5 × 10⁻⁴ || Acetic Acid | HCH₃CO₂ | 1.8 × 10⁻⁵ |### Base| **Name** | **Formula** | **Kb** ||----------------|--------------|-------------------|| Aniline | C₆H₅NH₂ | 4.3 × 10⁻¹⁰ || Methylamine | CH₃NH₂ | 4.4 × 10⁻⁴ |### Ranking Solutions by pHUse this data to rank the following solutions in order of increasing pH. In other words, select a '1' next to the solution that will have the lowest pH, a '2' next to the solution that will have the next lowest pH, and so on.| **Solution** | **pH** ||--------------------|-------------------|| **0.1 M C₆H₅NH₃Br** | ![dropdown](dropdown.png) || **0.1 M KCH₃CO₂** | ![dropdown](dropdown.png) || **0.1 M CH₃NH₃Cl** | ![dropdown](dropdown.png) || **0.1 M NaNO₂** | ![dropdown](dropdown.png) |Select the appropriate numbers (1-4) to correctly rank the solutions in terms of their pH levels:1. **Lowest pH**2. 3. 4. **Highest pH**---**Explanation of Ranking:**- Solutions with weak acids or their conjugate bases will have varying pH levels based on the given \(K_a\) and \(K_b\) values. - Typically, solutions with weak acids will have lower pH values, compared to their conjugate bases which will have higher pH values.- The provided \(K_a\) and \(K_b\) values give insight into the strength of the acids and bases, affecting their dissociation in water and hence their hydrogen ion concentration (p

Ka value of acetic acid = 1.8×10-5 Ka value of HNO2 = 4.5×10-4 Kb value of aniline = 4.3×10-10…

acid base K K₂b name formula name formula 10 nitrous acid HNO₂ 4.5 x 10 aniline |C HạNH, | 4.3 × 10 acetic acid HCH3CO₂ 1.8 x 10-5 methylamine CH3NH₂ 4.4×10 4 se this data to rank the following solutions in order of increasing pH. In other words, select a '1' next to the solution that will have the lowest pH, a '2' next to e solution that will have the next lowest pH, and so on. solution pH 0.1 M C6H5NH3Br ✓ choose one 1 (lowest) 0.1 M KCH3CO₂ 2 3 0.1 M CH3NH3CI 4 (highest) 0.1 M NaNO₂ choose one

acid base K K₂b name formula name formula 10 nitrous acid HNO₂ 4.5 x 10 aniline |C HạNH, | 4.3 × 10 acetic acid HCH3CO₂ 1.8 x 10-5 methylamine CH3NH₂ 4.4×10 4 se this data to rank the following solutions in order of increasing pH. In other words, select a '1' next to the solution that will have the lowest pH, a '2' next to e solution that will have the next lowest pH, and so on. solution pH 0.1 M C6H5NH3Br ✓ choose one 1 (lowest) 0.1 M KCH3CO₂ 2 3 0.1 M CH3NH3CI 4 (highest) 0.1 M NaNO₂ choose one

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

An acid (HA) dissociates as follows: HA → H* + A minus Arrow should be interpreted as an equilibrium arrow. The pH of the 0.81 M solution of HA is 1.46. What is the K, of HA? Express your answer as a decimal, not an exponent. No unit, please. Your Answer: Answer
K, for formic acid, HCOOH, is 1.80x10-4. Ka for acetylsalicylic acid (aspirin), HC,H,04, is 3.00x10-4. Ка Ką for nitrous acid, HNO2, is 4.50x10-4. What is the formula for the weakest conjugate base?
A solution is prepared at 25°C that is initially 0.081M in methylamine CH3NH2 , a weak base with =Kb×4.410−4 , and 0.070M in methylammonium chloride CH3NH3Cl . Calculate the pH of the solution. Round your answer to 2 decimal places.
Acids & Bases: Understanding Connections Between Descriptions of Weak Acid Dissociation In an aqueous solution of a certain acid the acid is 0.054% dissociated and the pH is 4.49. Calculate the acid dissociation constant Ka of the acid. Round your answer to 2 significant digits.
In order to determine the Ka of HXO₂, a scientist prepared a HXO₂ solution with a pH=3. If the scientist determined the Ka to be 1.2 X 10⁻⁵, what was the molarity of the acid he started with? (If you use any assumptions, be sure to check that they are valid!) Round the final answer to the three decimal places.
can you help me on the second problem please
Determine the pH, pOH, [H+], and [OH-] for each of the following problems. We may assume each acid base listed will 100# dissociate.
A solution is prepared that is initially 0.38M in ethylamine (C₂H5NH₂), a weak base, and 0.12M in ethylammonium chloride (C₂H5NH₂C1). Complete the reaction table below, so that you could use it to calculate the pH of this solution. Use x to stand for the unknown change in [OH-]. You can leave out the M symbol for molarity. initial change final [C,HẠNH,] 0 0 [C,HẠNH] 0 0 [OH-] 0 010 X Ś
Which of the following statements about acids is true? o The conjugate base of a strong acid will significantly raise the pH of water O For a polyprotic acid, the first H* dissociation has the smallest K, O The conjugate base of a weak acid is a stronger base than H2O All of the other statements are false For a weak acid, the equilibrium [H3O]* = initial [HA]
The pH of an aqueous solution at 25.0 °C that contains 3.98x10-9 M hydroxide ion is 6.50. TRUE OR FALSE
A solution is prepared at 25 °C that is initially 0.48 M in diethylamine ((C,H5)¸NH), a weak base with K,=1.3 × 10 -3 and 0.024M in diethylammonium bromide ((C,H5) NH,Br). Calculate the pH of the solution. Round your answer to 2 decimal places. pH olo Ar
Construct the expression for Ka for the weak acid, NH4+. 1 Based on the definition of Ka, drag the tiles into the numerator or denominator to construct the expression for the given acid. [H₂O] 2[OH-] [NH3]² NH.(aq) + H2O(I) = HO-(aq) + NH3(aq) [H3O+] 2[NH] [NH₂-]² Ka = [OH-] 2[NHs] [NH] 2[NH₂] [NH3] [H₂O]² [NH₂-] [H3O+]² 2[H₂O] [OH-]² ✓ RESET 2[H3O+] [NH