Calculate the pH of solution obtained by mixing 10 mL of 0.1 M HCl and 40 of 0.2 M H2SO4? [up to two decimal places]Correct answer is between '0.45,0.47'. Can you explain this answer?

Question

Answer

Calculating the pH of a Solution Obtained by Mixing HCl and H₂SO₄

Step 1: Calculate the moles of each acid

10 mL of 0.1 M HCl = 0.001 moles HCl

40 mL of 0.2 M H₂SO₄ = 0.008 moles H₂SO₄

Step 2: Determine the total volume of the solution

The total volume of the solution is 50 mL (10 mL + 40 mL)

Step 3: Calculate the concentration of H⁺ ions in the solution

HCl dissociates completely in water to form H⁺ and Cl^- ions. Therefore, the concentration of H⁺ ions contributed by HCl is 0.001 M.

H₂SO₄ partially dissociates in water to form H⁺ and HSO₄^- ions. The dissociation reaction is as follows: H₂SO₄ <-> H⁺ + HSO₄^-

The dissociation constant (K_a) for H₂SO₄ is 1.2 x 10^-2. Therefore, the equilibrium constant expression for the dissociation reaction is: K_a = [H⁺][HSO₄^-]/[H₂SO₄]

At equilibrium, the concentration of H⁺ ions is equal to the concentration of HSO₄^- ions, as HSO₄^- is the only other ion present in significant amounts. Therefore, [H⁺] = [HSO₄^-]

Substituting the values, we get: K_a = [H⁺]²/[H₂SO₄]

0.012 = [H⁺]²/(0.008 moles/0.05 L)

Solving for [H⁺], we get: [H⁺] = 0.12 M

Step 4: Calculate the pH

Answer

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