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A 50.0 mL solution of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH. What is the pH after 20.0 mL of base has been added? Ka CH3COOH = 1.8 x 10-5

A 50.0 mL solution of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH. What is the pH after 20.0 mL of base has been added? Ka CH3COOH = 1.8 x 10-5

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1. A 50.0 mL solution of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH. What is the pH after 20.0 mL of base has been added? Ka CH3COOH = 1.8 x 10-5

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(x)(0.02) 0.03 -5 1.8 × 10 3 0.02x = 5.4 x 10 -7 x = 2.7 x 10 -5 [H,O* (ag)] = x = 2.7 x 10 5 pH =- log[H3O*(ag)] - log(2.7 × 10 ³ ) = 4.57 Therefore, the pH after 20.0mL of base has been added is 4.57.
Transcribed Image Text:(x)(0.02) 0.03 -5 1.8 × 10 3 0.02x = 5.4 x 10 -7 x = 2.7 x 10 -5 [H,O* (ag)] = x = 2.7 x 10 5 pH =- log[H3O*(ag)] - log(2.7 × 10 ³ ) = 4.57 Therefore, the pH after 20.0mL of base has been added is 4.57.
1. CH;COOH CH;COOH + H¿0 ÷ H;0*+ CH;C00- (aq) ,+ NaOH(aq) → CH;COONa ag) + H,O CH;COOH (ag) NAOH (aq) CH;COONaa H,O (aq) B4 0.0075 0.003 After 0.0045 0.003 0.03 0.02 CH;COOH H,0 H,0* CH;C00¯ I 0.03М 0.02M C +x +x E 0.03 - x 0.02 +x -5 K. = 1.8 × 10 3 pH = ? [H,0 *][CH,CO0 | [CH;COOH ] Ка 1.8 × 10 -5 [x][0.02+x] [0.03-x] Rule of 100: [CH,COOH] Ка 0.03 1.8x10 = 1667 1667 » 100 Therefore, 0.03 –x = 0.03 Rule of 100: CH;C00 Ka 0.02 1.8x10 = 1111 1111 » 100 Therefore, 0.02 +x = 0.02
Transcribed Image Text:1. CH;COOH CH;COOH + H¿0 ÷ H;0*+ CH;C00- (aq) ,+ NaOH(aq) → CH;COONa ag) + H,O CH;COOH (ag) NAOH (aq) CH;COONaa H,O (aq) B4 0.0075 0.003 After 0.0045 0.003 0.03 0.02 CH;COOH H,0 H,0* CH;C00¯ I 0.03М 0.02M C +x +x E 0.03 - x 0.02 +x -5 K. = 1.8 × 10 3 pH = ? [H,0 *][CH,CO0 | [CH;COOH ] Ка 1.8 × 10 -5 [x][0.02+x] [0.03-x] Rule of 100: [CH,COOH] Ка 0.03 1.8x10 = 1667 1667 » 100 Therefore, 0.03 –x = 0.03 Rule of 100: CH;C00 Ka 0.02 1.8x10 = 1111 1111 » 100 Therefore, 0.02 +x = 0.02
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