Question

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 of solution. Determine the pH of the mixture. Determine the equilibrium concentrations of HF and HClO in the solution.

Answer 1

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.