The dissolution of CaCO3 in water is also affected by the equilibrium between the carbonate species. What is the solubility of CaCO3 in a water sample at 25°C initially without carbonate species (HCO3 and CO3²') at pH =10? The equilibrium constant for carbonate species is: HCO3¯ = H+ + CO3²- %3D [H*][Co,²-] K = 10-10.3 [HCO3¯] The solubility product of CaCO3 is Ksp = 106.4 at 25°C.

The dissolution of CaCO3 in water is also affected by the equilibrium between the carbonate species. What is the solubility of CaCO3 in a water sample at 25°C initially without carbonate species (HCO3 and CO3²') at pH =10? The equilibrium constant for carbonate species is: HCO3¯ = H+ + CO3²- %3D [H*][Co,²-] K = 10-10.3 [HCO3¯] The solubility product of CaCO3 is Ksp = 106.4 at 25°C.

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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Question

The dissolution of CaCO3 in water is also affected by the equilibrium between the carbonate species.
What is the solubility of CaCO3 in a water sample at 25°C initially without carbonate species (HCO3 and
CO3?) at pH =10? The equilibrium constant for carbonate species is:
HCO3 = H+ + Co,?-
[H*][C0;²-]
[HCO3¯]
3.
K =
10-10.3
The solubility product of CaCO3 is Ksp = 10-6.4 at 25°C.

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