Calculating [OH-] from an acid and salt.

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SUMMARY

The discussion focuses on calculating the hydroxide ion concentration ([OH-]) in a 0.5 M solution of potassium fluoride (KF). The dissociation of KF produces fluoride ions (F-) in a 1:1 ratio, resulting in [F-] = 0.5 M. The equilibrium expression using the acid dissociation constant (Ka) for hydrofluoric acid (HF), which is 7.2 × 10−4, is essential for determining the concentration of [OH-]. The reaction involves F- acting as a Brønsted base, reacting with water to form HF and OH-.

PREREQUISITES
  • Understanding of acid-base chemistry, particularly Brønsted-Lowry theory.
  • Familiarity with equilibrium expressions and ICE (Initial, Change, Equilibrium) tables.
  • Knowledge of the dissociation of salts in aqueous solutions.
  • Proficiency in calculating concentrations using equilibrium constants.
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  • Study the concept of Brønsted bases and their role in acid-base reactions.
  • Learn how to construct and interpret ICE tables for equilibrium calculations.
  • Research the relationship between Ka and Kb, particularly for weak acids and their conjugate bases.
  • Explore the calculation of pH and pOH in buffer solutions involving weak acids and their salts.
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Chemistry students, educators, and anyone involved in acid-base equilibrium calculations, particularly those working with weak acids and their salts.

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Homework Statement


Calculate the [OH-] concentration in a 0.5 M solution of potassium fluoride (KF). Ka for
HF is 7.2 × 10−4.

The Attempt at a Solution


I know that KF → K+ + F-, therefore having a 1:1 ratio, [F-] = 0.5M
,but from here I am unsure what to do with the F- concentration.

I know that Ka = ([H+][F-])/[HF] and normally I would set up an ICE chart, but I would have too many variables in this case. (Not knowing the concentration of [HF] as well as the "change")
 
Last edited:
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F- + H2O <-> HF + OH-

Or, to put it differently - F- is a Bronsted base.
 

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