Answer:
* ![pH=11.0](https://tex.z-dn.net/?f=pH%3D11.0)
* ![[HClO]_{eq}=9.32x10^{-4}M](https://tex.z-dn.net/?f=%5BHClO%5D_%7Beq%7D%3D9.32x10%5E%7B-4%7DM)
* ![[HF]_{eq}=5.07x10^{-6}M](https://tex.z-dn.net/?f=%5BHF%5D_%7Beq%7D%3D5.07x10%5E%7B-6%7DM)
Explanation:
Hello,
At first, the molarities of NaClO and KF are computed as shown below:
![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](https://tex.z-dn.net/?f=M_%7BNaClO%7D%3D%5Cfrac%7B51.2g%2A%5Cfrac%7B1mol%7D%7B74.45g%7D%20%7D%7B0.250L%7D%3D2.75M%20%5C%5C%5C%5CM_%7BKF%7D%3D%5Cfrac%7B25.3g%2A%5Cfrac%7B1mol%7D%7B58.1g%7D%20%7D%7B0.250L%7D%20%3D1.74M)
Now, since both NaClO and KF are ionic, one proposes the dissociation reactions as:
![NaClO\rightarrow Na^++ClO^-\\KF\rightarrow K^++F^-](https://tex.z-dn.net/?f=NaClO%5Crightarrow%20Na%5E%2B%2BClO%5E-%5C%5CKF%5Crightarrow%20K%5E%2B%2BF%5E-)
In such a way, by writing the formation of HClO and HF respectively, due to the action of water, we obtain:
![ClO^-+H_2O\rightleftharpoons HClO+OH^-\\F^-+H_2O\rightleftharpoons HF+OH^-\\](https://tex.z-dn.net/?f=ClO%5E-%2BH_2O%5Crightleftharpoons%20HClO%2BOH%5E-%5C%5CF%5E-%2BH_2O%5Crightleftharpoons%20HF%2BOH%5E-%5C%5C)
Now, the presence of OH⁻ immediately implies that Kb for both NaClO and KF must be known as follows:
![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}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7BKw%7D%7BKa%7D%20%5C%5CKb_%7BNaClO%7D%3D%5Cfrac%7B1x10%5E%7B-14%7D%7D%7B10%5E%7B-7.50%7D%7D%3D3.16x10%5E%7B-7%7D%5C%5CKb_%7BKF%7D%3D%5Cfrac%7B1x10%5E%7B-14%7D%7D%7B10%5E%7B-3.17%7D%7D%3D1.48x10%5E%7B-11%7D)
In such a way, the law of mass action for each case is:
![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}}](https://tex.z-dn.net/?f=Kb_%7BNaClO%7D%3D3.16x10%5E%7B-7%7D%3D%5Cfrac%7B%5BHClO%5D%5BOH%5D%5E-%7D%7B%5BClO%5E-%5D%7D%3D%5Cfrac%7B%28x_%7BClO%7D%29%28x_%7BClO%7D%29%7D%7B2.75-x_%7BClO%7D%7D%5C%5CKb_%7BKF%7D%3D1.48x10%5E%7B-11%7D%3D%5Cfrac%7B%5BHF%5D%5BOH%5D%5E-%7D%7B%5BF%5E-%5D%7D%3D%5Cfrac%7B%28x_%7BF%7D%29%28x_%7BF%7D%29%7D%7B1.74-x_%7BF%7D%7D)
Now, by solving for the change
for both the ClO⁻ and F⁻ equilibriums, we obtain:
![x_{ClO}=9.32x10^{-4}M\\x_{F}=5.07x10^{-6}M](https://tex.z-dn.net/?f=x_%7BClO%7D%3D9.32x10%5E%7B-4%7DM%5C%5Cx_%7BF%7D%3D5.07x10%5E%7B-6%7DM)
Such results equal the concentrations of OH⁻ in each equilibrium, thus, the total concentration of OH⁻ result:
![[OH^-]_{tot}=9.32x10^{-4}M+5.07x10^{-6}M=9.37x10^{-4}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D_%7Btot%7D%3D9.32x10%5E%7B-4%7DM%2B5.07x10%5E%7B-6%7DM%3D9.37x10%5E%7B-4%7DM)
With which the pOH is:
![pOH=-log(9.37x10^{-4})=3.03](https://tex.z-dn.net/?f=pOH%3D-log%289.37x10%5E%7B-4%7D%29%3D3.03)
And the pH:
![pH=14-pOH=11.0](https://tex.z-dn.net/?f=pH%3D14-pOH%3D11.0)
In addition, the equilibrium concentrations of HClO and HF equals the change
for each equilibrium as:
![[HClO]_{eq}=x_{ClO}=9.32x10^{-4}M](https://tex.z-dn.net/?f=%5BHClO%5D_%7Beq%7D%3Dx_%7BClO%7D%3D9.32x10%5E%7B-4%7DM)
![[HF]_{eq}=x_{F}=5.07x10^{-6}M](https://tex.z-dn.net/?f=%5BHF%5D_%7Beq%7D%3Dx_%7BF%7D%3D5.07x10%5E%7B-6%7DM)
Best regards.