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Consider the titration of 50.0 mL of 2.0 M HNO₂ (Ka=4.0x10⁻⁴) with 1.0 M KOH.

Calculate the pH before any base is added.

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Final answer:

To find the pH of a 2.0 M HNO2 solution before any base is added, we can use the provided Ka value to set up an ICE table, solve for the concentration of H+ ions, and then calculate the pH. After solving, we find that the pH of the solution is approximately 2.05.

Step-by-step explanation:

To calculate the pH before any base is added, we first recognize that HNO₂ is a weak acid, meaning it does not completely dissociate in water. We can use the given Ka value of 4.0x10⁻⁴ to determine the concentration of hydrogen ions (H⁺) produced in the solution.

For the reaction:

HNO₂ → H⁺ + NO₂⁻

We set up an ICE table (Initial, Change, Equilibrium) to find the equilibrium concentrations:

Initial: [HNO₂] = 2.0 M, [H⁺] = 0, [NO₂⁻] = 0

Change: [HNO₂] decreases by x, [H⁺] increases by x, [NO₂⁻] increases by x

Equilibrium: [HNO₂] = 2.0 - x, [H⁺] = x, [NO₂⁻] = x

Using the equation Ka = [H⁺][NO₂⁻]/[HNO₂], we have:

4.0x10⁻⁴ = (x)(x) / (2.0 - x) ≈ x² / 2.0 (since x is small relative to 2.0 M)

Solving for x, we get x = √(4.0x10⁻⁴ * 2.0) = √(8.0x10⁻⁴) ≈ 8.94x10⁻⁳ M

Now, we can calculate the pH:

pH = -log([H⁺]) = -log(8.94x10⁻⁳) ≈ 2.05

Therefore, the pH of the 2.0 M HNO₂ solution before any base is added is approximately 2.05.

User Dmytro Shevchenko
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