# 1442answerstest3 - Answers 1442 Test 3 Chapters 18 and 19 1...

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Unformatted text preview: Answers: 1442 Test 3 Chapters 18 and 19 1. pH = -log[H + ] 3.4 = - log[H + ]-3.4 = log[H + ] [H + ] = 3.98 X 10-4 ( enter -3.4 and hit 2nd log on the calculator) 2. pOH = -log[OH] = -log[1.2 X 10-6 ] = 5.9 pH + pOH = 14 pH + 5.92 = 14 pH = 8.08 3. HBr → H + + Br- pH = -long[H + ] = -log[.08] = 1.1 .08 .08 .08 4. Ca(OH) 2 → Ca 2+ + 2 OH- pOH = -log[.1]= 1, pH = 13 .05 .05 .1 5. R HOCl ↔ H + + OCl- .5 0 -x x x .5-x x x Ka = [x 2 ] / [.5-x] 3.5 X 10-8 = x 2 / .5 ( ignore the x because it is so small. If Ka is 10-4 or smaller( 10-5 ...) you can probably ignore x) x = 1.32 X 10-4 , Since x = H + the pH is 3.88 6. R C 6 H 5 NH 2 + H 2 O ↔ C 6 H 5 NH 3 + + OH- .3 0 -x x x .3-x x x Kb = [x 2 ] / [.3-x] 4.2 X 10-10 = x 2 / .3 ( ignore the x) x = 1.12 X 10-5 , Since x = OH- , pOH = 4.95, pH = 9.05 7. R HX ↔ H + + X - I .1 0 C-.003 +.003 +.003 * E .097 .003 .003 Ka = [.003] 2 / [.097] = 9.28 X 10-5 Since the acid is 3% (.03) ionized, to get the change you multiply .1 X .03 = .003 3% ionized or dissociated means 3% of the initial amount breaks up into products. 8. R HX ↔ H + + X- I .1 0 0 C-.01 .01 .01* E .09 .01 .01 Ka = [.01] 2 / .09 = 1.1 X 10-3 Since the pH is 2: solve for 2 = -log[H + ] which gives H + = .01 9. R HF ↔ H + + F- I .04* 0 .06 C -x x x E .04 - x x .06 + x Ka = [x][.06 + x] / [.04-x] 7.2 X 10-4 = [x][.06] / [ .04] Ignoring both x's....
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