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24 June, 17:55

A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 * 10-10. What is the pOH after 30.5 mL of NaOH is added? The total volume is 25.0 mL + 30.5 mL = 55.5 mL.

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  1. 24 June, 20:25
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    The pOH = 1.83

    Explanation:

    Step 1: Data given

    volume of the sample = 25.0 mL

    Molarity of hydrocyanic acid = 0.150 M

    Molarity of NaOH = 0.150 M

    Ka of hydrocyanic acid = 4.9 * 10^-10

    Step 2: The balanced equation

    HCN + NaOH → NaCN + H2O

    Step 3: Calculate the number of moles hydrocyanic acid (HCN)

    Moles HCN = molarity * volume

    Moles HCN = 0.150 M * 0.0250 L

    Moles HCN = 0.00375 moles

    Step 3: Calculate moles NaOH

    Moles NaOH = 0.150 M * 0.0305 L

    Moles NaOH = 0.004575 moles

    Step 4: Calculate the limiting reactant

    0.00375 moles HCN will react with 0.004575 moles NaOH

    HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH

    Step 5: Calculate molarity of NaOH

    Molarity NaOH = moles NaOH / volume

    Molarity NaOH = 0.000825 moles / 0.0555 L

    Molarity NaOH = 0.0149 M

    Step 6: Calculate pOH

    pOH = - log [OH-]

    pOH = - log (0.0149)

    pOH = 1.83

    The pOH = 1.83
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