17.43 A 20.0-mL sample of 0.200 M HBr solution is titrated with 0.200 M NaOH sol

Question

17.43 A 20.0-mL sample of 0.200 M HBr solution is titrated with 0.200 M NaOH solution. Calculate the pH of the solu tion after the following volumes of base have been added: (a) 15.0 mL, (b) 19.9 mL, (c) 20.0 mL, (d) 20.1 mL, (e) 35.0 mL 17.44 A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HCIO4 solution. Calculate the pH after the following vol- umes of acid have been added: (a) 20.0 mL, (b) 23.0 mL, (c) 24.0 mL, (d) 25.0 mL, (e) 30.0 mL. A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is ti- trated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added: (a) 0 mL, (b) 17.5 mL, (c) 34.5 mL, (d) 35.0 mL, (e) 35.5 mL, (f) 50.0 mL 17.45 17.46 Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 MHCI. Calculate the pH after the following volumes of titrant have been added: (a) 0 mL, (b) 20.0 mL, (c) 59.0 mL, (d) 60.0 mL, (e) 61.0 mL, (f) 65.0 mL. 17.47 Calculate the pH at the equivalence point for titrating 0.200 M solutions of each of the following bases with 0.200 M HBr: (a) sodium hydroxide (NaOH), (b) hydroxylamine (NH2OH), (c) aniline (CGHsNH2) Calculate the pH at the equivalence point in titrating 0.100 M solutions of each of the following with 0.080 M NaOH: (a) hydrobromic acid (HBr), (b) chlorous acid (HCIO2), (c) benzoic acid (C6HsCOOH) 17.48

Explanation / Answer

17.43

HBr + OH- --> Br- + H2O

Strong acid-strong base titration

(a) 15 ml NaOH added

initial moles HBr = 0.2 M x 20 ml = 4 mmol

moles NaOH added = 0.2 M x 15 ml = 3 mmol

[HBr] remained = 1 mmol/35 ml = 0.03 M

pH = -log[H+]

      = -log(0.03)

      = 1.54

(b) 15 ml NaOH added

initial moles HBr = 0.2 M x 20 ml = 4 mmol

moles NaOH added = 0.2 M x 19.9 ml = 3.98 mmol

[HBr] remained = 0.02 mmol/39.9 ml = 5.01 x 10^-4 M

pH = -log[H+]

      = -log(5.01 x 10^-4)

      = 3.30

(c) 20 ml NaOH added

all of acid is neutralized

pH = 7.0

(d) 20.1 ml NaOH added

excess [OH-] = 0.2 M x 0.1 ml/40.1 ml = 0.0005 M

pOH = -log[OH-]

        = -log(0.0005)

        = 3.30

pH = 14 - pOH

     = 14 - 3.30

     = 10.7

(e) 35 ml NaOH added

excess [OH-] = 0.2 M x 15 ml/55 ml = 0.054 M

pOH = -log[OH-]

        = -log(0.054)

        = 1.26

pH = 14 - pOH

     = 14 - 1.26

     = 12.74

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17.45

CH3COOH + OH- ---> CH3COO- + H2O

(a) 0 ml NaOH added

CH3COOH <==> CH3COO- + H+

x amount dissociated

Ka = 1.8 x 10^-5 = x^2/0.150

x = [H+] = 1.64 x 10^-3 M

pH = -log[H+] = 2.78

(b) 17.5 ml NaOH added

initial moles CH3COOH = 0.150 M x 35 ml = 5.25 mmol

moles NaOH added = 0.150 M x 17.5 ml = 2.625 mmol

[CH3COOH] remined = [CH3COO-] formed

pH = pKa = 4.76

(c) 34.5 ml NaOH added

initial moles CH3COOH = 0.150 M x 35 ml = 5.25 mmol

moles NaOH added = 0.150 M x 34.5 ml = 5.175 mmol

(CH3COOH) remined = 0.075 mmol

(CH3COO-) formed = 5.175

pH = 4.76 + log(5.175/0.075) = 6.60

(d) 35 ml NaOH added

[CH3COO-] formed = 0.150 M x 35 ml/70 ml = 0.075 M

CH3COO- + H2O <==> CH3COOH + OH-

let x amount hydrolyzed

Kb = 5.55 x 10^-10 = x^2/0.075

x = [OH-] = 6.45 x 10^-6 M

pOH = -log[OH-] = 5.19

pH = 14 - pOH = 8.81

(e) 35.5 ml NaOH added

excess [OH-] = 0.150 M x 0.5 ml/70.5 ml = 1.06 x 10^-3 M

pOH = -log(1.06 x 10^-3) = 2.97

pH = 14 - 2.97 = 11.03

(f) 50 ml NaOH added

excess [OH-] = 0.150 M x 15 ml/85 ml = 0.026 M

pOH = -log(0.026) = 1.58

pH = 14 - 1.58 = 12.42

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