A solution is prepared by mixing 499 mL of 0.19 M Tris·Base and 592 mL of 0.37 M Tris·Acid. What is the pH of this solution? [Ka(Tris·Acid) = 8.5 × 10-9] Round your answer to two places past the decimal.

I can do basic pH questions but I am at a loss on this one. Please help.

Use the Henderson-Hasselbalch equation.

To find the pH of the solution, we need to calculate the concentration of H+ ions in the solution.

First, let's calculate the moles of Tris·Base and Tris·Acid in the solution:

Moles of Tris·Base = Volume (in liters) x Concentration = 499 mL x 0.19 M = 0.09481 moles
Moles of Tris·Acid = Volume (in liters) x Concentration = 592 mL x 0.37 M = 0.21864 moles

Next, we need to calculate the initial concentration of Tris·Acid. Since Tris·Acid acts as a weak acid, it will partially dissociate in water. Let's assume x is the amount of Tris·Acid that reacts with water.

Tris·Acid (initial) = Moles of Tris·Acid - x
Water (initial) = Moles of Tris·Acid + x

The dissociation reaction of Tris·Acid in water is:
Tris·Acid + H2O ⇌ Tris·Base + H+

From the balanced equation, we can see that 1 mole of Tris·Acid reacts to form 1 mole of Tris·Base and 1 mole of H+. Therefore, the moles of Tris·Acid that react (x) will also be equal to the moles of H+ ions formed.

Now, let's set up an expression for the equilibrium constant (Ka) using the given Ka value and the equilibrium concentrations of the species:

Ka = [Tris·Base] [H+] / [Tris·Acid]

Since we have the initial concentrations, we can substitute the initial concentrations for each species:

Ka = x * x / (0.21864 - x)

The concentration of H+ ions (y) will be equal to x. Now, we can substitute y for x in the equation above:

Ka = y^2 / (0.21864 - y)

Simplifying this equation, we have:

8.5 × 10^(-9) = y^2 / (0.21864 - y)

Rearranging the equation:

8.5 × 10^(-9) (0.21864 - y) = y^2

Multiplying both sides:

8.5 × 10^(-9) (0.21864) - 8.5 × 10^(-9) y = y^2

Now, let's rearrange this equation to make it a quadratic equation:

y^2 + 8.5 × 10^(-9) y - 1.8456 × 10^(-9) = 0

Solving this quadratic equation, we find two possible values for y, which represent the concentration of H+ ions. One value is positive and one is negative. Since the concentration of H+ ions cannot be negative, we disregard the negative value.

Using the quadratic formula, we find:

y ≈ 1.28 × 10^(-5) M

Finally, to find the pH, we can use the formula:

pH = -log[H+]

pH = -log(1.28 × 10^(-5))

pH ≈ 4.89

Therefore, the pH of the solution is approximately 4.89.