HW2_29

# HW2_29 - 3 NH 4 1 NH 3 H 1 let x =[H 1 produced[NH 4 1[NH...

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Homework due 2/29 Calculate the pH of a 0.25 M NH 4 NO 3 solution. (K b for NH 3 = 1.8x10 -5 ) This is a typical salt hydrolysis problem in which one must determine the pH of a solution of the salt of a strong acid (HNO 3 ) and a weak base (NH 3 ). First the salt must be ionized then the pH of the weak acid NH 4 1+ solution can be determined. One gets the K a for the weak acid from the K b for NH 3 its conjugate base according to the equation K a K b = K w . NH 4 NO 3 NH 4 1+ + NO 3 1- (spectator ion since from the strong acid HNO
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Unformatted text preview: 3 ) NH 4 1+ NH 3 + H 1+ let x = [H 1+ ] produced [NH 4 1+ ] [NH 3 ] [H 1+ ] initial 0.25 (from NH 4 NO 3 ) change-x x x equil 0.25 -x x x K a = K w /K b = 1.0x10-14 /1.8x10-5 = 5.6x10-10 (NH 4 1+ ) (NH 3 ) [NH 3 ][H 1+ ] (x)(x) K a = ------------------- = ----------------- = 5.6x10-10 [NH 4 1+ ] (0.25 - x) ignoring the subtracted x and solving for x gives x = [H 1+ ] = 1.18x10-5 pH = -log [H 1+ ] = 4.92...
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