
Identify each of the following as an acid or a base: (10.1)
- H2SO4
- RbOH
- Ca(OH)2
- HI

To identify: Each of the following as an acid or base.
- H2SO4
- RbOH
- Ca(OH)2
- HI
Answer to Problem 10.71UTC
Solution:
- H2SO4 − Acid.
- RbOH − Base.
- Ca(OH)2 − Base.
- HI − Acid.
a. Given: H2SO4
Explanation of Solution
According to Arrhenius concept, substance that dissociates in water to give H+ ions or increases the concentration of H+ ions is said to be acidic in nature.
So, H2SO4is an acid as it gives H+ ions in aqueous solution.
b. Given: RbOH
According to Arrhenius concept, substance that dissociates in water to give OH- ions or increases the concentration of OH- ions is said to be basic in nature.
So, RbOH is a base as it gives OH- ions in aqueous solution.
c. Given: Ca(OH)2
According to Arrhenius concept, substance that dissociates in water to give OH- ions or increases the concentration of OH- ions is said to be basic in nature.
So, Ca(OH)2is a base as it gives OH- ions in aqueous solution.
d. Given: HI
According to Arrhenius concept, substance that dissociates in water to give H+ ions or increases the concentration of H+ ions is said to be acidic in nature.
So, HI is an acid as it gives H+ ions in aqueous solution.
Thus H2SO4 and HI are acids whereas RbOH and Ca(OH)2 are bases.
Want to see more full solutions like this?
Chapter 10 Solutions
Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)
Additional Science Textbook Solutions
Biology: Life on Earth with Physiology (11th Edition)
Cosmic Perspective Fundamentals
Applications and Investigations in Earth Science (9th Edition)
Human Physiology: An Integrated Approach (8th Edition)
Chemistry: The Central Science (14th Edition)
Microbiology with Diseases by Body System (5th Edition)
- 2' P17E.6 The oxidation of NO to NO 2 2 NO(g) + O2(g) → 2NO2(g), proceeds by the following mechanism: NO + NO → N₂O₂ k₁ N2O2 NO NO K = N2O2 + O2 → NO2 + NO₂ Ко Verify that application of the steady-state approximation to the intermediate N2O2 results in the rate law d[NO₂] _ 2kk₁[NO][O₂] = dt k+k₁₂[O₂]arrow_forwardPLEASE ANSWER BOTH i) and ii) !!!!arrow_forwardE17E.2(a) The following mechanism has been proposed for the decomposition of ozone in the atmosphere: 03 → 0₂+0 k₁ O₁₂+0 → 03 K →> 2 k₁ Show that if the third step is rate limiting, then the rate law for the decomposition of O3 is second-order in O3 and of order −1 in O̟.arrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning
- Introduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage Learning



