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8 May, 17:14

A buffer is prepared by adding 139 mL of 0.39 M NH3 to 169 mL of 0.19 M NH4NO3. What is the pH of the final solution? (Assume the value of Kw is 1.0 ✕ 10-14. Enter an unrounded value.)

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  1. 8 May, 17:28
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    pH = 9.48

    Explanation:

    We have first to realize that NH₃ is a weak base:

    NH₃ + H₂O ⇔ NH₄⁺ + OH⁻ Kb = 1.8 x 10⁻⁵

    and we are adding this weak base to a solution of NH₄NO₃ which being a salt dissociates 100 % in water.

    Effectively what we have here is a buffer of a weak base and its conjugate acid. Therefore, we need the Henderson-Hasselbach formula for weak bases given by:

    pOH = pKb + log ([ conjugate acid ] / [ weak base ]

    mol NH₃ = 0.139 L x 0.39 M = 0.054 mol

    mol NH₄⁺ = 0.169 L x 0.19 M = 0.032 mol

    Now we have all the information required to calculate the pOH (Note that we dont have to calculate the concentrations since in the formula they are a ratio and the volume will cancel out)

    pOH = - log (1.8 x 10⁻⁵) + log (0.032/0.054) = 4.52

    pOH + pH = 14 ⇒ pH = 14 - 4.52 = 9.48

    The solution is basic which agrees with NH₃ being a weak base.
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