Question #105912
A buffered solution contains 0.25 M NH3 (Kb = 1.8 x 10-5) and 0.40 M NH4Cl. a. Calculate the pH of this solution. b. Calculate the pH of the solution that results when 0.10 mole of gaseous HCl is added to 1.0 L of the buffered solution from part a.
1
Expert's answer
2020-03-25T12:46:52-0400

pOH=pKb+log[[NH4+][NH4OH]]pOH=pK_b+log[\frac{[NH_4^+]}{[NH_4OH]} ]

pOH=log(1.8×105)+log[0.40.25]pOH=-log(1.8×10^{-5})+log[\frac{0.4}{0.25} ]

pOH=log(1.8)+5+log(1.6)pOH=-log(1.8)+5+log(1.6)

pOH=5+log(1.61.8)pOH=5+log(\frac{1.6}{1.8})

pOH=5+log(16)log(18)pOH=5+log(16)-log(18)

pOH=5+4log(2)log(9×2)pOH=5+4log(2)-log(9×2)

pOH=5+4log(2)log(2)2log(3)pOH=5+4log(2)-log(2)-2log(3)

pOH=5+3×0.3012×0.477pOH=5+3×0.301-2×0.477

pOH=4.949pOH=4.949

pH=9.051pH=9.051

(b)pOH=pKb+log[[NH4+][NH4OH]]pOH=pK_b+log[\frac{[NH_4^+]}{[NH_4OH]} ]

pOH=log(1.8×105)+log[0.50.15]pOH=-log(1.8×10^{-5})+log[\frac{0.5}{0.15} ]

pOH=log(1.8×105)+log[0.50.15]pOH=-log(1.8×10^{-5})+log[\frac{0.5}{0.15} ]

pOH=5log(1.8)+log[103]pOH=5-log(1.8)+log[\frac{10}{3} ]

pOH=5log(1.8)+log[10]log(3)pOH=5-log(1.8)+log[10 ]-log(3)

pOH=5log(18)+2log(3)pOH=5-log(18)+2-log(3)

pOH=5log(2)+23log(3)pOH=5-log(2)+2-3log(3)

pOH=5.27pOH=5.27

pH=8.73pH=8.73


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