Equal volume of 0.06M NA2SO4 (aq) the and 0.06 M KHSO4(aq) the are mixed. I Calculate pH of resultant solution. Given pKa(HSO4-) = 2*10^(-2)

To find the pH of the solution formed by mixing equal volumes of 0.06 M Na2SO4 and 0.06 M KHSO4, we need to consider the dissociation of these compounds in water and how they interact with each other. Let's break this down step by step.

Understanding the Components

When Na2SO4 is dissolved in water, it dissociates completely into sodium ions (Na⁺) and sulfate ions (SO₄^2-). The sulfate ion is a very weak base and does not significantly affect the pH. On the other hand, KHSO4 dissociates into potassium ions (K⁺) and hydrogen sulfate ions (HSO₄^-), which can act as a weak acid.

Key Reactions

In this case, the relevant equilibrium reaction for HSO₄^- can be represented as follows:

• HSO₄^- ⇌ H⁺ + SO₄^2-

Calculating the pH

Since both solutions are mixed in equal volumes, the concentrations of the ions will remain the same after mixing. Therefore, we still have 0.06 M HSO₄^- in the solution. The dissociation constant (Ka) for HSO₄^- can be calculated from the given pKa:

pKa = -log(Ka) implies that:

Ka = 10^-pKa = 10^-2 = 0.01

Setting Up the Equilibrium Expression

For the dissociation of HSO₄^-, we can set up the equilibrium expression:

• Ka = [H⁺][SO₄^2-] / [HSO₄^-]

Let x be the concentration of H⁺ produced at equilibrium. Initially, we have:

• [HSO₄^-] = 0.06 M
• [H⁺] = 0
• [SO₄^2-] = 0

At equilibrium, we have:

• [HSO₄^-] = 0.06 - x
• [H⁺] = x
• [SO₄^2-] = x

Substituting into the Ka Expression

Now we can substitute these values into the Ka expression:

0.01 = (x)(x) / (0.06 - x)

Assuming x is small compared to 0.06, we can simplify this to:

0.01 ≈ (x²) / 0.06

Solving for x

Now, rearranging gives:

x² = 0.01 * 0.06 = 0.0006

x = √0.0006 ≈ 0.0245 M

Finding the pH

The concentration of H⁺ ions in the solution is approximately 0.0245 M. To find the pH, we use the formula:

pH = -log[H⁺]

pH = -log(0.0245) ≈ 1.61

Thus, the pH of the resultant solution after mixing equal volumes of 0.06 M Na2SO4 and 0.06 M KHSO4 is approximately 1.61. This indicates that the solution is acidic, primarily due to the presence of the HSO₄^- ions.