Answer:
* [tex]pH=11.0[/tex]
* [tex][HClO]_{eq}=9.32x10^{-4}M[/tex]
* [tex][HF]_{eq}=5.07x10^{-6}M[/tex]
Explanation:
Hello,
At first, the molarities of NaClO and KF are computed as shown below:
[tex]M_{NaClO}=\frac{51.2g*\frac{1mol}{74.45g} }{0.250L}=2.75M \\\\M_{KF}=\frac{25.3g*\frac{1mol}{58.1g} }{0.250L} =1.74M[/tex]
Now, since both NaClO and KF are ionic, one proposes the dissociation reactions as:
[tex]NaClO\rightarrow Na^++ClO^-\\KF\rightarrow K^++F^-[/tex]
In such a way, by writing the formation of HClO and HF respectively, due to the action of water, we obtain:
[tex]ClO^-+H_2O\rightleftharpoons HClO+OH^-\\F^-+H_2O\rightleftharpoons HF+OH^-\\[/tex]
Now, the presence of OH⁻ immediately implies that Kb for both NaClO and KF must be known as follows:
[tex]Kb=\frac{Kw}{Ka} \\Kb_{NaClO}=\frac{1x10^{-14}}{10^{-7.50}}=3.16x10^{-7}\\Kb_{KF}=\frac{1x10^{-14}}{10^{-3.17}}=1.48x10^{-11}[/tex]
In such a way, the law of mass action for each case is:
[tex]Kb_{NaClO}=3.16x10^{-7}=\frac{[HClO][OH]^-}{[ClO^-]}=\frac{(x_{ClO})(x_{ClO})}{2.75-x_{ClO}}\\Kb_{KF}=1.48x10^{-11}=\frac{[HF][OH]^-}{[F^-]}=\frac{(x_{F})(x_{F})}{1.74-x_{F}}[/tex]
Now, by solving for the change [tex]x[/tex] for both the ClO⁻ and F⁻ equilibriums, we obtain:
[tex]x_{ClO}=9.32x10^{-4}M\\x_{F}=5.07x10^{-6}M[/tex]
Such results equal the concentrations of OH⁻ in each equilibrium, thus, the total concentration of OH⁻ result:
[tex][OH^-]_{tot}=9.32x10^{-4}M+5.07x10^{-6}M=9.37x10^{-4}M[/tex]
With which the pOH is:
[tex]pOH=-log(9.37x10^{-4})=3.03[/tex]
And the pH:
[tex]pH=14-pOH=11.0[/tex]
In addition, the equilibrium concentrations of HClO and HF equals the change [tex]x[/tex] for each equilibrium as:
[tex][HClO]_{eq}=x_{ClO}=9.32x10^{-4}M[/tex]
[tex][HF]_{eq}=x_{F}=5.07x10^{-6}M[/tex]
Best regards.
