NH 3 aq H aq dissociation in water Since H is a stronger acid than NH 4 NH 4

# Nh 3 aq h aq dissociation in water since h is a

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NH 3 (aq) + H + (aq) (dissociation in water) Since H + is a stronger acid than NH 4 + , NH 4 + cannot donate its proton to water to reform NH 3
[H + ] = = 9.1 10 -3 M 1.0×10 mol −3 (0.060 + 0.050) L × pH 2.04 = b. Adding 0.10M NaOH to 50.0mL of 0.10M CH 3 COOH: Point 1: Before NaOH was added Equation: HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O 2 - (aq) [HC 2 H 3 O 2 ] [H + ] [C 2 H 3 O 2 - ] I 0.10 M 0 0 C – x + x + x E 0.10 – x x x .8 ; x H ] 1.3 M ; K a = 1 × 10 −5 = x 2 (0.10 − x ) x 2 (0.10) = [ + = × 10 −3 pH .89 = 2 Point 2: Before equivalence point: 10.0mL of NaOH is added to the solution Equation: HC 2 H 3 O 2 (aq) + OH - (aq) H 2 O(l) + C 2 H 3 O 2 - (aq) (neutralization) OH - HC 2 H 3 O 2 C 2 H 3 O 2 - I = 1.0 10 -3 mol 1 L 0.10 mol ×0.0100 L × = 5.0 10 -3 mol 1 L 0.10 mol ×0.0500 L × 0 mol C - 1.0 10 -3 × - 1.0 10 -3 × + 1.0 10 -3 × E 0 mol 4.0 10 -3 mol × 1.0 10 -3 mol × Equation: C 2 H 3 O 2 - (aq) + H 2 O(l) HC 2 H 3 O 2 (aq) + OH - (aq) (dissociation of conjugate base) [C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] [OH - ] I =0.016667 1.0×10 mol −3 (0.010 + 0.050) L =0.06667 4.0×10 mol −3 (0.010 + 0.050) L 0 C – x + x + x E 0.016667 – x 0.06667 + x x ; x OH ] 1.4 M ; pOH 9.86 K b = 1.8 ×10 −5 1.0 ×10 −14 = (0.016667− x ) x (0.06667 + x ) (0.016667) x (0.06667) = [ = × 10 −10 = pH 4.00 9.86 4.14 = 1 =
Point 3: At halfway of the equivalent point: 25.0mL of NaOH is added to the solution Equation: HC 2 H 3 O 2 (aq) + OH - (aq) H 2 O(l) + C 2 H 3 O 2 - (aq) (neutralization) OH - HC 2 H 3 O 2 C 2 H 3 O 2 - I = 2.5 10 -3 mol 1 L 0.10 mol ×0.0250 L × = 5.0 10 -3 mol 1 L 0.10 mol ×0.0500 L × 0 mol C - 2.5 10 -3 × - 2.5 10 -3 × + 2.5 10 -3 × E 0 mol 2.5 10 -3 mol × 2.5 10 -3 mol × Equation: HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O 2 - (aq) (dissociation of weak acid) [HC 2 H 3 O 2 ] [C 2 H 3 O 2 - ] [H + ] I =0.03333 2.5×10 mol −3 (0.025 + 0.050) L =0.03333 2.5×10 mol −3 (0.025 + 0.050) L 0 C – x + x + x E 0.033333 – x 0.033333 + x x .8 ; x H ] .8 M ; pH 4.74 K a = 1 × 10 −5 = (0.33333− x ) x (0.03333 + x ) (0.03333) x (0.03333) = [ + = 1 × 10 −5 = Point 4: At equivalence point: 50.0mL of NaOH is added to the solution Equation: HC 2 H 3 O 2 (aq) + OH - (aq) H 2 O(l) + C 2 H 3 O 2 - (aq) (neutralization) OH - HC 2 H 3 O 2 C 2 H 3 O 2 - I = 5.0 10 -3 mol 1 L 0.10 mol ×0.0500 L × = 5.0 10 -3 mol 1 L 0.10 mol ×0.0500 L × 0 mol C - 5.0 10 -3 × - 5.0 10 -3 × + 5.0 10 -3 × E 0 mol 0 mol 5.0 10 -3 mol × Equation: C 2 H 3 O 2 - (aq) + H 2 O(l) HC 2 H 3 O 2 (aq) + OH - (aq) (dissociation of conjugate base) [C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] [OH - ] I = 0.050 5.0×10 mol −3 (0.050 + 0.050) L 0 0 C – x + x + x E 0.050 – x x x
; x H ] .3 M ; pOH 5.28 K b = 1.8 ×10 −5 1.0 ×10 −14 = x 2 (0.050 − x ) x 2 (0.050) = [ + = 5 × 10 −6 = H 14.00 5.28 8.72 p = = Point 5: After equivalence point: 60.0mL of NaOH is added to the solution Equation: HC 2 H 3 O 2 (aq) + OH - (aq) H 2 O(l) + C 2 H 3 O 2 - (aq) (neutralization) OH - HC 2 H 3 O 2 C 2 H 3 O 2 - I = 6.0 10 -3 mol 1 L 0.10 mol ×0.0500 L × = 5.0 10 -3 mol 1 L 0.10 mol ×0.0500 L × 0 mol C - 5.0 10 -3 × - 5.0 10 -3 × + 5.0 10 -3 × E 1.0 10 -3 mol × 0 mol 5.0 10 -3 mol × Equation: C 2 H 3 O 2 - (aq) + H 2 O(l) HC 2 H 3 O 2 (aq) + OH - (aq) (dissociation of conjugate base) Since OH - is a very strong base compared to acetate ions; the acetate ions cannot take a proton from water to reform acetic acid. [OH - ] = = 9.1 10 -3 M 1.0×10 mol −3 (0.060 + 0.050) L × OH 2.04 ; pH 14.00 2.04 11.96 p = = = V. ANALYSIS QUESTIONS:

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• Fall '15
• Neethling,Melanie
• Chemistry, pH, Equivalence point, 10-3 mol