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sattari [20]
3 years ago
8

1. 51.2 g of NaClO (pKa of HClO is 7.50) and 25.3 g of KF (pKa of of HF is 3.17) were combined in enough water to make 0.250 L o

f solution. Determine the pH of the mixture. Determine the equilibrium concentrations of HF and HClO in the solution.
Chemistry
1 answer:
Andrews [41]3 years ago
6 0

Answer:

* pH=11.0

* [HClO]_{eq}=9.32x10^{-4}M

* [HF]_{eq}=5.07x10^{-6}M

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

Now, since both NaClO and KF are ionic, one proposes the dissociation reactions as:

NaClO\rightarrow Na^++ClO^-\\KF\rightarrow K^++F^-

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^-\\

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}

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}}

Now, by solving for the change x for both the ClO⁻ and F⁻ equilibriums, we obtain:

x_{ClO}=9.32x10^{-4}M\\x_{F}=5.07x10^{-6}M

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

With which the pOH is:

pOH=-log(9.37x10^{-4})=3.03

And the pH:

pH=14-pOH=11.0

In addition, the equilibrium concentrations of HClO and HF equals the change x for each equilibrium as:

[HClO]_{eq}=x_{ClO}=9.32x10^{-4}M

[HF]_{eq}=x_{F}=5.07x10^{-6}M

Best regards.

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For cube C: L=3 cm, W=3 cm

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2) Surface area of each cube

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