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25 July, 11:34

What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = pKa? Ka = 1.8*10-5 Quantity = mol Submit b What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 4.00? Ka = 1.8*10-5 Quantity = mol c What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 5.00? Ka = 1.8*10-5 Quantity = mol

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  1. 25 July, 11:47
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    a) We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2

    b) We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2

    c) We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2

    Explanation:

    a) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = pKa?

    Step 1: Data given

    Volume of HC2H3O2 = 1.0 L

    Molarity of HC2H3O2 = 1.8 M

    Ka = 1.8*10^-5

    ph = pK = - log (1.8*10^-5) = 4.74

    Step 2:

    Use the Henderson-Hasselbalch equation.

    pH = pKa + log (A-/HA)

    4.74 = 4.74 + log (A-/HA)

    0 = log (A-/HA)

    A-/HA = 1

    Consider X = moles of NaOH added (and moles of A - formed)

    Remaining moles of HA = 1.8 - X

    moles of A - = X

    HA = 1.8 - X

    X / (1.8-X) = 1

    X = 0.9

    We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2

    To control we can do the following equation:

    4.74 = 4.74 + log (0.9/0.9) = 4.74

    b) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 4.00?

    Step 1: Data given

    Volume of HC2H3O2 = 1.0 L

    Molarity of HC2H3O2 = 1.8 M

    Ka = 1.8*10^-5

    ph = 4

    Step 2:

    Use the Henderson-Hasselbalch equation.

    pH = pKa + log (A-/HA)

    4 = 4.74 + log (A-/HA)

    -0.74 = log (A-/HA)

    A-/HA = 0.182

    Consider X = moles of NaOH added (and moles of A - formed)

    Remaining moles of HA = 1.8 - X

    moles of A - = X

    HA = 1.8 - X

    X / (1.8-X) = 0.182

    X = 0.277

    We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2

    To control we can do the following equation:

    4 = 4.74 + log (0.277/1.523)

    c) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 5.00

    Step 1: Data given

    Volume of HC2H3O2 = 1.0 L

    Molarity of HC2H3O2 = 1.8 M

    Ka = 1.8*10^-5

    ph = 5

    Step 2:

    Use the Henderson-Hasselbalch equation.

    pH = pKa + log (A-/HA)

    5 = 4.74 + log (A-/HA)

    0.26 = log (A-/HA)

    A-/HA = 1.82

    Consider X = moles of NaOH added (and moles of A - formed)

    Remaining moles of HA = 1.8 - X

    moles of A - = X

    HA = 1.8 - X

    X / (1.8-X) = 1.82

    X = 1.16

    We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2

    To control we can do the following equation:

    5 = 4.74 + log (1.16/0.64) = 5
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