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18 September, 07:34

An analytical chemist is titrating 242.5mL of a 1.200M solution of hydrazoic acid HN3 with a 0.3400M solution of NaOH. The pKa of hydrazoic acid is 4.72.

Calculate the pH of the acid solution after the chemist has added 1006. mL of the NaOH solution to it.

Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added. Round your answer to 2 decimal places.

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  1. 18 September, 10:23
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    The pH of the solution is 12.61

    Explanation:

    Step 1: Data given

    Concentration of hydrazoic acid solution = 1.200 M

    Volume of the solution = 242.5 mL = 0.2425 L

    Concentration of NaOH solution = 0.3400 M

    The pKa of hydrazoic acid is 4.72

    Step 2: The balanced equation

    NaOH + HN3 → NaN3 + H2O

    Step 3: Calculate moles hydrazoic acid

    Moles hydrazoic acid (HN3) = concentration * volume

    Moles hydrazoic acid = 1.200 M * 0.2425 L

    Moles hydrazoic acid = 0.291 moles

    Step 4: Calculate moles NaOH

    Moles NaOH = 0.3400 M * 1.006 L

    Moles NaOH = 0.342 moles

    Step 5: Determine the limiting reactant

    HN3 is the limiting reactant. It will completely be consumed (0.291 moles)

    NaOH is in excess. There reacts 0.291 moles. There will remain 0.342 - 0.291 = 0.051 moles NaOH

    Step 6: Calculate total volume

    Total volume = 242.5 mL + 1006 mL = 1248.5 mL = 1.2485 L

    Step 7: Calculate concentration of NaOH

    concentration NaOH = 0.051 moles / 1.2485 L

    concentration NaOH = 0.0408 M

    Step 8: Calculate pOH

    pOH = - log [OH-] = - log (0.0408)

    pOH = 1.39

    Step 9: Calculate pH

    pH + pOH = 14

    pH = 14 - pOH

    pH = 14 - 1.39

    pH = 12.61

    The pH of the solution is 12.61
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