10) Determine the Ka of a 0.86 M solution of Hydrofluoric acid, HF that is 1.72%ionized at 25 °C.Write the equation for the dissociation of an acid and define an expression for KaDraw an ICE tableWhat other information in the problem above can help you determine the value of x?Use that information to solve for the value of x and substitute back into the ICE tableto determine equilibrium concentration of products and reactantsSubstitute the equilibrium concentration of reactants and products into the Kaexpression to determine the value of Ka

Given, Concentration of HF = 0.86 M % ionisation = 1.72 %

10) Determine the Ka of a 0.86 M solution of Hydrofluoric acid, HF that is 1.72% ionized at 25 °C. Write the equation for the dissociation of an acid and define an expression for Ka Draw an ICE table

10) Determine the Ka of a 0.86 M solution of Hydrofluoric acid, HF that is 1.72% ionized at 25 °C. Write the equation for the dissociation of an acid and define an expression for Ka Draw an ICE table

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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Bronsted Lowry Base In Inorganic Chemistry

Bronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.

Question

10) Determine the Ka of a 0.86 M solution of Hydrofluoric acid, HF that is 1.72%
ionized at 25 °C.
Write the equation for the dissociation of an acid and define an expression for Ka
Draw an ICE table
What other information in the problem above can help you determine the value of x?
Use that information to solve for the value of x and substitute back into the ICE table
to determine equilibrium concentration of products and reactants
Substitute the equilibrium concentration of reactants and products into the Ka
expression to determine the value of Ka

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