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.

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Explanation:

Hello,

At first, the molarities of NaClO and KF are computed as shown below:

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

In such a way, by writing the formation of HClO and HF respectively, due to the action of water, we obtain:

Now, the presence of OH⁻ immediately implies that Kb for both NaClO and KF must be known as follows:

In such a way, the law of mass action for each case is:

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

Such results equal the concentrations of OH⁻ in each equilibrium, thus, the total concentration of OH⁻ result:

With which the pOH is:

And the pH:

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

Best regards.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.