Yes nh 3 nh 4 c l system n h 3 aq h 2 o aq 10m 10m kb

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Unformatted text preview: its to maintain pH in a solution? YES!!! NH 3 + NH 4 C l system N H 3 (aq) +H 2 O (aq) 1.0M 1.0M Kb NH 4 + (aq) +O H - (aq) 1.0M + 1.0M Kw → H 2 O (l) ← H + ( aq ) Add H+ [ NH 3 ] [ N H 4 + ][O H − ] − [OH ] = K b Kb = [ NH 4 + ] [ NH 3 ] NH 4 + (aq) Add Add OH- Ka = K w Kb [NH 4 + ] pOH = pK b + log [NH3 ] [A - ] N H 3 (aq) +H 3 O + (aq) pH = pK a + log [HA] Calculate Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl 0.30 buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH 4 + (aq) K = K K NH 3 (aq) +H 3O + (aq) a [A - ] pH = pK a + log [HA] 0.08 × 0.36 =0.029 nf cf 0.028 0 .028 0.10 b pK a = − log(10 −14 1.8 × 10 −5 ) = 9.25 pH = 9.25 + log(0.30 0.36) = 9.17 NH + (aq) + OH - (a...
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