Can someone solve these 2 problems for me?
1. When 0.0017 moles of a weak acid are dissolved in enough water to make 0.500 L, the pH of the resulting solution is 5.78. Calculate Ka.
2. A solution is made by dissolving 15.5 g of NaOH in approximately 450 mL of water in a volumetric flask. The solution becomes quite warm, but after it is allowed to return to room temperature, water is added to total 500 mL of solution. Calculate the pH of the final solution. Report pH to 2 decimal places.
1.Let's call the weak acid HA, then (HA) in mols/L = 0.0017 moles/0.500L = 0.0034 M. Then
..................HA ==> H^+ + A^-
I................0.0034..........0.....0
C..................-x.............x......x
E............0.0034 - x.......x......x
pH = -log (H^+) = 5.78. Solve for (H^+). It will be approximately 1E-6 and that will be x but you should use the more exact answer.
Substitute the E line (equilibrium) in to the Ka expression and solve for Ka. You will need to use the quadratic equation. You may want to think about correcting for the H^+ from H2O but that won't amount to more than about 5% error and I wouldn't try to correct for that. I don't think it's worth the effort since the Ka values are not known any better than that.
2. On both 1 and 2 please post your work if you get stuck.
For 2 you have 15.5 g NaOH in 500 mL solution which is 15.5 g/40 g/mol = 0.3875 mols/0.500 L = 0.775 M
So NaOH is strong base, it ionizes completely so (OH^-) will be 0.775 M.
Convert that to pOH then convert to pH using
pH + pOH = pKw = 14. You know pKw and pOH so solve for pH.
pOH = about 0.1 so pH = about 13.9
Sure! I can help you solve these problems and also explain the steps to get the answers. Let's start with the first problem.
1. When 0.0017 moles of a weak acid are dissolved in enough water to make 0.500 L, the pH of the resulting solution is 5.78. We are asked to calculate Ka.
To solve this problem, we need to use the equilibrium expression for the ionization of the weak acid, which is:
Ka = [H+][A-] / [HA]
In this equation, [H+] represents the concentration of hydronium ions, [A-] represents the concentration of the conjugate base, and [HA] represents the initial concentration of the weak acid.
Since we are given the pH of the resulting solution, we can calculate the concentration of hydronium ions ([H+]) using the formula:
[H+] = 10^(-pH)
[H+] = 10^(-5.78)
Now that we know the concentration of hydronium ions, we can use the fact that the weak acid is dissociating in a 1:1 ratio with the conjugate base (A-) to determine the concentration of the conjugate base ([A-]).
[A-] = [H+]
Finally, we can use the given moles of weak acid and the volume of the solution to calculate the initial concentration of the weak acid ([HA]).
[HA] = moles of weak acid / volume of solution
Plugging all these values into the equilibrium expression for Ka, we can calculate the Ka.
Now let's move on to the second problem.
2. In this problem, we are asked to calculate the pH of a final solution after adding water to a solution containing NaOH.
To solve this problem, we need to consider the dissociation of NaOH in water. NaOH dissociates completely into Na+ and OH- ions.
The hydroxide ion (OH-) reacts with water to form hydronium ions (H3O+). The equilibrium expression for this reaction is:
OH- + H2O ⟷ H2O + H3O+
Since NaOH is a strong base and dissociates completely, we can assume that the concentration of hydroxide ions ([OH-]) in the solution is equal to the initial concentration of NaOH.
To calculate the concentration of hydroxide ions ([OH-]), we need to convert the given mass of NaOH to moles, and then divide by the final volume of the solution.
Once we have the concentration of hydroxide ions ([OH-]), we can use the fact that water auto-dissociates into hydronium ions (H3O+) and hydroxide ions (OH-).
At room temperature, the concentration of hydronium ions ([H3O+]) and hydroxide ions ([OH-]) in pure water is 1.0 x 10^(-7) M.
Since [H3O+] × [OH-] = Kw = 1.0 x 10^(-14) at room temperature, we can use this relationship to calculate the concentration of hydronium ions ([H3O+]) in the final solution.
[H3O+] = Kw / [OH-]
Finally, we can use the formula:
pH = -log[H3O+]
to calculate the pH of the final solution.
Remember, when performing calculations, it's important to keep track of units and conversions to ensure accurate results.