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Solution: Solve an equilibrium problem (using an ICE table) to calculate the pH of each
Chapter 16, Problem 33(choose chapter or problem)
Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution.
a. 0.15 M HF
b. 0.15 M NaF
c. a mixture that is 0.15 M in HF and 0.15 M in NaF
Questions & Answers
QUESTION:
Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution.
a. 0.15 M HF
b. 0.15 M NaF
c. a mixture that is 0.15 M in HF and 0.15 M in NaF
ANSWER:Step 1 of 5
Using an ICE table we need to calculate the pH of the given solutions.
In the ICE table I stands for the initial concentrations, C stands for the changes in concentration, and E stands for the equilibrium concentrations. Use multiples of x based on the stoichiometric coefficients in the balanced equilibrium equation to represent the changes in concentration in terms of x .Reactants decrease, so their changes are given negative values. Products increase, so their changes are given positive values. Keep in mind that the concentrations of liquids are never included in an ICE table
Negative logarithm of hydrogen ion concentration is called as \(\mathrm{pH}\).
Mathematically
\(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]\)
The negative logarithm of the hydroxide \(\left(\mathrm{OH}^{-}\right)\) ion concentration of a solution is called as \(\mathrm{pOH}\)
\(\mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]\)
The values \(\mathrm{pH}\) and \(\mathrm{pOH}\) re related to each other as
\(\mathrm{pK}_{\mathrm{w}}=\mathrm{pH}+\mathrm{pOH}\)