**Interactive Table Completion Activity****Instructions:**Complete the table below. Ensure each of your entries has the correct number of significant digits. Assume the temperature is 25 °C.| Conjugate Acid | | Conjugate Base | ||----------------|----------|-----------------|----------|| Formula | \( K_a \) | Formula | \( K_b \) || HF | \( 6.8 \times 10^{-4} \) | [Box] | [Box] || [Box] | [Box] | HS\(^{-}\) | \( 1.8 \times 10^{-7} \) || [Box] | [Box] | C\(_4\)H\(_8\)ONH | \( 2.1 \times 10^{-6} \) |**Note:**- Each placeholder [Box] represents an area for input within the table, where users are expected to fill in the missing information based on their understanding of acid-base equilibrium.- The activity includes an interactive section with icons for submitting, resetting the response, and asking for help.**Icons:**- A checkmark (✓) for submission.- A circular arrow for resetting the input.- A question mark (?) for guidance or hints.This activity is designed to help students practice calculating the missing values of dissociation constants for acid-base pairs.

According to Lowery and Bronsted, an acid is defined as a substance which has a tendency to donate a proton to any other substance and a base is a substance which has a tendency to accept the proton from any other substance.A) Given molecule: Conjugate acid =HF Chemical species that differ in composition only by a proton are called a conjugate pair. thus Every acid has a conjugate base, which is formed when the acid donates a proton. Given molecule HF is an conjugate acid which will donate a proton to form a conjugate base. write the reaction: HF → H+ + F-Conjugate Conjugate acid base Therefore, the conjugate base ofHF is F-. Given Ka of the acid = 6.8×10-4 Relation between acid dissociation constant Ka and base dissociation constant Kb is given by: Ka × Kb = KWWhere,Ka = Acid dissociation constant Kb = Base dissociation constant KW =Water dissociation constant Substitute, 6.8×10-4 Ka , 10-14 Kw in the expression: 6.8 ×10-4 ×Kb = 10-14Kb = 10-146.8 ×10-4 ≈ 1.5 × 10-11 Therefore, Kb of the conjugate base is 1.5×10-11.B) Given molecule: Conjugate base = HS- Chemical species that differ in composition only by a proton are called a conjugate pair. thus Every base has a conjugate acid, which is formed when the base, accepts a proton. Given molecule HS- is an conjugate base which will accepts a proton to form a conjugate acid. write the reaction: HS- + H+ → H2SConjugate Conjugate base acid Therefore, the conjugate acid of HS- is H2S. Given Kb of the base = 1.8×10-7 Relation between acid dissociation constant Ka and base dissociation constant Kb is given by: Ka × Kb = KWWhere,Ka = Acid dissociation constant Kb = Base dissociation constant KW =Water dissociation constant Substitute, Kb 1.8×10-7, 10-14 Kw in the expression: 1.8 ×10-7 ×Ka = 10-14Ka = 10-141.8 ×10-7 ≈ 5.6 ×10-8 Therefore, Ka of the conjugate acid is 5.6×10-8.C) Given molecule: Conjugate base = C4H8ONH Chemical species that differ in composition only by a proton are called a conjugate pair. Thus Every base has a conjugate acid, which is formed when the base, accepts a proton. Given molecule C4H8ONH is an conjugate base which will accepts a proton to form a conjugate acid. write the reaction: C4H8ONH + H+ → C4H8ONH2+Conjugate Conjugate base acid Therefore, the conjugate acid of C4H8ONH is C4H8ONH2+. Given Kb of the base = 2.1×10-6 Relation between acid dissociation constant Ka and base dissociation constant Kb is given by: Ka × Kb = KWWhere,Ka = Acid dissociation constant Kb = Base dissociation constant KW =Water dissociation constant Substitute, Kb 2.1×10-6, 10-14 Kw in the expression: 2.1 ×10-6 ×Ka = 10-14Ka = 10-142.1 ×10-6 ≈ 4.8 ×10-9 Therefore, Ka of the conjugate acid is 4.8×10-9.

Answered: conjugate acid conjugate base Ka…

Science

Chemistry

conjugate acid conjugate base Ka formula formula HF 6.8 x 10 HS 1.8 x 10 9- C,H,ONH 2.1 x 10

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...

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