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22 July, 05:36

How many grams of dry nh4cl need to be added to 2.50 l of a 0.800 m solution of ammonia, nh3, to prepare a buffer solution that has a ph of 8.53? kb for ammonia is 1.8*10-5. express your answer with the appropriate units?

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  1. 22 July, 05:40
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    The correct answer is 574.59 grams.

    Explanation:

    Based on the given information, the number of moles of NH₃ will be,

    = 2.50 L * 0.800 mol/L

    = 2 mol

    The given pH of a buffer is 8.53

    pH + pOH = 14.00

    pOH = 14.00 - pH

    pOH = 14.00 - 8.53

    pOH = 5.47

    The Kb of ammonia given is 1.8 * 10^-5. Now pKb = - logKb,

    = - log (1.8 * 10⁻⁵)

    = 5.00 - log 1.8

    = 5.00 - 0.26

    = 4.74

    Based on Henderson equation:

    pOH = pKb + log ([salt]/[base])

    pOH = pKb + [NH₄⁺]/[NH₃]

    5.47 = 4.74 + log ([NH₄⁺]/[NH₃])

    log ([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73

    [NH₄⁺]/[NH₃] = 10^0.73 = 5.37

    [NH₄⁺ = 5.37 * 2 mol = 10.74 mol

    Now the mass of dry ammonium chloride required is,

    mass of NH₄Cl = 10.74 mol * 53.5 g/mol

    = 574.59 grams.
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