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26 August, 07:42

An analytical chemist is titrating 242.5 mL of a 1.200 M solution of hydrazoic acid HN3 with a 0.3400 M 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. 26 August, 07:57
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    The pH of the solution is 12.61

    Explanation:

    Step 1: Data given

    Molarity of hydrazoic acid solution = 1.200 M

    Volume of solution = 242.5 mL

    Molarity of NaOH solution = 0.3400 M

    pKa = 4.72

    Step 2: The balanced equation

    HN3 + NaOH → NaN3 + H2O

    Step 3: Calculate moles hydrazoic acid

    Moles hydrazoic acid = molarity * volume

    Moles hydra

    zoic 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: Calculate the limiting reactant

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

    NaOH is in excess. There reacts 0.29 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 molarity of NaOH

    Molarity NaOH = 0.051 moles / 1.2485 L

    Molarity NaOH = 0.0408 M

    Step 8: Calculate pOH

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

    pOH = 1.39

    Step 9: Calculate pH

    pH = 14 - pOH

    pH = 14 - 1.39

    pH = 12.61

    The pH of the solution is 12.61
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