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23 January, 15:25

A 1.00 l solution contains 24.52 g of nitrous acid, hno2. what mass of sodium nitrite, nano2, should be added to it to make a buffer with a ph of 2.96

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  1. 23 January, 16:42
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    When moles of HNO2 = mass/molar mass

    =24.52 g / 47g/mol

    = 0.52 mol

    [HNO2] = moles / volume

    = 0.52 mol * 1 L

    = 0.52 M

    by using H-H equation:

    PH = Pka + ㏒[salt / acid]

    when we have Ka = 4 x 10^-4

    ∴Pka = - ㏒Ka

    = - ㏒ (4x 10^-4)

    = 3.4

    by substitution in H-H equation:

    2.96 = 3.4 + ㏒[NaNO2/0.52]

    ∴[NaNO2] = 0.189 M

    when moles NaNO2 = 0.189 M * 1 L = 0.189 mol

    ∴Mass of NaNO2 = moles NaNO2 * molar mass

    = 0.189 mol * 68.99g/mol

    = 13 g
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