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Wyatt's Notes — IB

Acids and Bases -- Diagnostic Tests

Acids and Bases — Diagnostic Tests

Unit Tests

UT-1: Weak Acid pH and Percentage Ionisation

Question: Ethanoic acid has Ka=1.74×10−5 mol dm−3K_a = 1.74 \times 10^{-5}\ \text{mol dm}^{-3} at 25 ∘C25\ ^\circ\text{C}. Calculate the pH of a 0.150 mol dm−30.150\ \text{mol dm}^{-3} solution and the percentage ionisation. State the approximation used and verify it is valid.

Solution: CH3COOH⇌CH3COO−+H+\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+

Ka=[CH3COO−][H+][CH3COOH]K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]}

Let x=[H+]=[CH3COO−]x = [\text{H}^+] = [\text{CH}_3\text{COO}^-]. Then [CH3COOH]≈0.150−x[\text{CH}_3\text{COOH}] \approx 0.150 - x.

Approximation: since KaK_a is very small, x≪0.150x \ll 0.150So 0.150−x≈0.1500.150 - x \approx 0.150.

1.74×10−5=x20.1501.74 \times 10^{-5} = \frac{x^2}{0.150}

x2=2.61×10−6x^2 = 2.61 \times 10^{-6}

x=1.616×10−3 mol dm−3x = 1.616 \times 10^{-3}\ \text{mol dm}^{-3}

pH=−log⁡10(1.616×10−3)=2.79\text{pH} = -\log_{10}(1.616 \times 10^{-3}) = 2.79

Verification: x/0.150=0.0108=1.08%<5%x/0.150 = 0.0108 = 1.08\% \lt 5\%. The approximation is valid.

Percentage ionisation: 1.616×10−30.150×100=1.08%\frac{1.616 \times 10^{-3}}{0.150} \times 100 = 1.08\%.

UT-2: Buffer Solution pH Calculation

Question: A buffer is prepared by adding 0.100 mol0.100\ \text{mol} of sodium ethanoate to 100 cm3100\ \text{cm}^3 of 1.00 mol dm−31.00\ \text{mol dm}^{-3} ethanoic acid (Ka=1.74×10−5K_a = 1.74 \times 10^{-5}). Calculate the pH of the buffer. Then calculate the change in pH when 5.00 cm35.00\ \text{cm}^3 of 2.00 mol dm−32.00\ \text{mol dm}^{-3} HCl is added to 50.0 cm350.0\ \text{cm}^3 of this buffer.

Solution:

Buffer pH using Henderson-Hasselbalch:

Moles of CH3COOH\text{CH}_3\text{COOH}: 1.00×0.100=0.100 mol1.00 \times 0.100 = 0.100\ \text{mol} Moles of CH3COO−\text{CH}_3\text{COO}^-: 0.100 mol0.100\ \text{mol}

pH=pKa+log⁡[A−][HA]=−log⁡(1.74×10−5)+log⁡0.1000.100=4.76+0=4.76\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]} = -\log(1.74 \times 10^{-5}) + \log\frac{0.100}{0.100} = 4.76 + 0 = 4.76

Adding HCl to 50.0 cm350.0\ \text{cm}^3 of buffer:

Moles CH3COOH\text{CH}_3\text{COOH} in 50 cm350\ \text{cm}^3: 0.100×50100=0.0500 mol0.100 \times \frac{50}{100} = 0.0500\ \text{mol} Moles CH3COO−\text{CH}_3\text{COO}^- in 50 cm350\ \text{cm}^3: 0.100×50100=0.0500 mol0.100 \times \frac{50}{100} = 0.0500\ \text{mol} Moles HCl added: 2.00×0.00500=0.0100 mol2.00 \times 0.00500 = 0.0100\ \text{mol}

HCl reacts with CH3COO−\text{CH}_3\text{COO}^-: CH3COO−+H+→CH3COOH\text{CH}_3\text{COO}^- + \text{H}^+ \to \text{CH}_3\text{COOH}

New moles: CH3COOH=0.0500+0.0100=0.0600 mol\text{CH}_3\text{COOH} = 0.0500 + 0.0100 = 0.0600\ \text{mol}, CH3COO−=0.0500−0.0100=0.0400 mol\text{CH}_3\text{COO}^- = 0.0500 - 0.0100 = 0.0400\ \text{mol}.

pH=4.76+log⁡0.04000.0600=4.76+log⁡(0.667)=4.76−0.176=4.58\text{pH} = 4.76 + \log\frac{0.0400}{0.0600} = 4.76 + \log(0.667) = 4.76 - 0.176 = 4.58

PH change: 4.76−4.58=0.184.76 - 4.58 = 0.18. The buffer resists the pH change effectively.

UT-3: Salt Hydrolysis pH

Question: Calculate the pH of a 0.0500 mol dm−30.0500\ \text{mol dm}^{-3} solution of ammonium chloride (NH4Cl\text{NH}_4\text{Cl}). Kb(NH3)=1.78×10−5K_b(\text{NH}_3) = 1.78 \times 10^{-5}.

Solution: NH4+\text{NH}_4^+ is the conjugate acid of NH3\text{NH}_3. It hydrolyses in water:

NH4++H2O⇌NH3+H3O+\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+

Ka(NH4+)=KwKb(NH3)=1.00×10−141.78×10−5=5.62×10−10K_a(\text{NH}_4^+) = \frac{K_w}{K_b(\text{NH}_3)} = \frac{1.00 \times 10^{-14}}{1.78 \times 10^{-5}} = 5.62 \times 10^{-10}

Ka=[NH3][H+][NH4+]=x20.0500K_a = \frac{[\text{NH}_3][\text{H}^+]}{[\text{NH}_4^+]} = \frac{x^2}{0.0500}

x2=5.62×10−10×0.0500=2.81×10−11x^2 = 5.62 \times 10^{-10} \times 0.0500 = 2.81 \times 10^{-11}

x=5.30×10−6 mol dm−3x = 5.30 \times 10^{-6}\ \text{mol dm}^{-3}

pH=−log⁡(5.30×10−6)=5.28\text{pH} = -\log(5.30 \times 10^{-6}) = 5.28

The solution is acidic, as expected for a salt of a weak base and a strong acid.

Integration Tests

IT-1: Titration Curve Analysis (with Measurement and Data Processing)

Question: 25.0 cm325.0\ \text{cm}^3 of 0.100 mol dm−30.100\ \text{mol dm}^{-3} ethanoic acid (Ka=1.74×10−5K_a = 1.74 \times 10^{-5}) is titrated with 0.100 mol dm−30.100\ \text{mol dm}^{-3} NaOH. Calculate: (a) the initial pH, (b) the pH at the half-equivalence point, (c) the pH at the equivalence point, (d) the volume of NaOH at the equivalence point. Identify a suitable indicator.

Solution:

(a) Initial pH: [H+]=Ka×c=1.74×10−5×0.100=1.32×10−3[\text{H}^+] = \sqrt{K_a \times c} = \sqrt{1.74 \times 10^{-5} \times 0.100} = 1.32 \times 10^{-3}. pH=2.88\text{pH} = 2.88.

(b) At half-equivalence, [HA]=[A−][\text{HA}] = [\text{A}^-]So pH=pKa=4.76\text{pH} = \text{p}K_a = 4.76.

(c) At equivalence, all CH3COOH\text{CH}_3\text{COOH} has been converted to CH3COO−\text{CH}_3\text{COO}^-. The concentration of CH3COO−\text{CH}_3\text{COO}^- is:

Total volume =25.0+25.0=50.0 cm3= 25.0 + 25.0 = 50.0\ \text{cm}^3. [CH3COO−]=0.100×25.050.0=0.0500 mol dm−3[\text{CH}_3\text{COO}^-] = \frac{0.100 \times 25.0}{50.0} = 0.0500\ \text{mol dm}^{-3}.

Kb=KwKa=1.00×10−141.74×10−5=5.75×10−10K_b = \frac{K_w}{K_a} = \frac{1.00 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.75 \times 10^{-10}.

[OH−]=Kb×0.0500=2.875×10−11=5.36×10−6[\text{OH}^-] = \sqrt{K_b \times 0.0500} = \sqrt{2.875 \times 10^{-11}} = 5.36 \times 10^{-6}.

pOH=5.27\text{pOH} = 5.27, pH=14.00−5.27=8.73\text{pH} = 14.00 - 5.27 = 8.73.

(d) Volume: 0.100×25.00.100=25.0 cm3\frac{0.100 \times 25.0}{0.100} = 25.0\ \text{cm}^3.

Suitable indicator: phenolphthalein (pH range 8.28.2—10.010.0), since the equivalence pH (8.73) falls within this range.

IT-2: Polyprotic Acid and Equilibrium (with Equilibrium)

Question: Carbonic acid (H2CO3\text{H}_2\text{CO}_3) is a diprotic acid with Ka1=4.3×10−7K_{a1} = 4.3 \times 10^{-7} and Ka2=4.8×10−11K_{a2} = 4.8 \times 10^{-11}. Calculate the pH of a 0.0200 mol dm−30.0200\ \text{mol dm}^{-3} solution. Justify why you can ignore the second dissociation.

Solution: First dissociation: H2CO3⇌H++HCO3−\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-

Ka1=[H+][HCO3−][H2CO3]=x20.0200=4.3×10−7K_{a1} = \frac{[\text{H}^+][\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} = \frac{x^2}{0.0200} = 4.3 \times 10^{-7}

x2=8.6×10−9x^2 = 8.6 \times 10^{-9}

x=9.27×10−5 mol dm−3x = 9.27 \times 10^{-5}\ \text{mol dm}^{-3}

pH=−log⁡(9.27×10−5)=4.03\text{pH} = -\log(9.27 \times 10^{-5}) = 4.03

Justification for ignoring second dissociation: Ka2≪Ka1K_{a2} \ll K_{a1} (ratio ≈10−4\approx 10^{-4}). The second dissociation (HCO3−⇌H++CO32−\text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-}) produces a negligible additional [H+][\text{H}^+] compared to the first. The additional [H+][\text{H}^+] from the second step would be approximately Ka2≈4.8×10−11K_{a2} \approx 4.8 \times 10^{-11}Which is orders of magnitude smaller than 9.27×10−59.27 \times 10^{-5}. The second dissociation contributes less than 0.00005%0.00005\% of the total [H+][\text{H}^+].

Verification of approximation: x/0.0200=0.46%<5%x/0.0200 = 0.46\% \lt 5\%. Valid.

IT-3: Buffer Capacity and Stoichiometry (with Stoichiometry)

Question: A student prepares a buffer by mixing 50.0 cm350.0\ \text{cm}^3 of 0.200 mol dm−30.200\ \text{mol dm}^{-3} NaOH\text{NaOH} with 75.0 cm375.0\ \text{cm}^3 of 0.200 mol dm−30.200\ \text{mol dm}^{-3} ethanoic acid (Ka=1.74×10−5K_a = 1.74 \times 10^{-5}). Calculate the pH. Then determine the maximum volume of 0.500 mol dm−30.500\ \text{mol dm}^{-3} HCl that can be added to 40.0 cm340.0\ \text{cm}^3 of this buffer before the pH drops below 4.00.

Solution:

Moles NaOH=0.200×0.0500=0.0100 mol\text{NaOH} = 0.200 \times 0.0500 = 0.0100\ \text{mol} Moles CH3COOH=0.200×0.0750=0.0150 mol\text{CH}_3\text{COOH} = 0.200 \times 0.0750 = 0.0150\ \text{mol}

NaOH\text{NaOH} neutralises some CH3COOH\text{CH}_3\text{COOH}:

CH3COOH\text{CH}_3\text{COOH} remaining: 0.0150−0.0100=0.00500 mol0.0150 - 0.0100 = 0.00500\ \text{mol} CH3COO−\text{CH}_3\text{COO}^- formed: 0.0100 mol0.0100\ \text{mol}

pH=4.76+log⁡0.01000.00500=4.76+0.301=5.06\text{pH} = 4.76 + \log\frac{0.0100}{0.00500} = 4.76 + 0.301 = 5.06

For the buffer capacity question, in 40.0 cm340.0\ \text{cm}^3:

Moles CH3COOH=0.00500×40125=0.00160 mol\text{CH}_3\text{COOH} = 0.00500 \times \frac{40}{125} = 0.00160\ \text{mol} Moles CH3COO−=0.0100×40125=0.00320 mol\text{CH}_3\text{COO}^- = 0.0100 \times \frac{40}{125} = 0.00320\ \text{mol}

After adding v cm3v\ \text{cm}^3 of HCl (0.500 mol dm−30.500\ \text{mol dm}^{-3}): moles HCl =0.500v/1000=5.00×10−4v= 0.500v/1000 = 5.00 \times 10^{-4}v

CH3COOHnew=0.00160+5.00×10−4v\text{CH}_3\text{COOH}_{\text{new}} = 0.00160 + 5.00 \times 10^{-4}v CH3COOnew−=0.00320−5.00×10−4v\text{CH}_3\text{COO}^-_{\text{new}} = 0.00320 - 5.00 \times 10^{-4}v

pH=4.76+log⁡0.00320−5.00×10−4v0.00160+5.00×10−4v=4.00\text{pH} = 4.76 + \log\frac{0.00320 - 5.00 \times 10^{-4}v}{0.00160 + 5.00 \times 10^{-4}v} = 4.00

4.00−4.76=log⁡0.00320−5.00×10−4v0.00160+5.00×10−4v4.00 - 4.76 = \log\frac{0.00320 - 5.00 \times 10^{-4}v}{0.00160 + 5.00 \times 10^{-4}v}

10−0.76=0.174=0.00320−5.00×10−4v0.00160+5.00×10−4v10^{-0.76} = 0.174 = \frac{0.00320 - 5.00 \times 10^{-4}v}{0.00160 + 5.00 \times 10^{-4}v}

0.174(0.00160+5.00×10−4v)=0.00320−5.00×10−4v0.174(0.00160 + 5.00 \times 10^{-4}v) = 0.00320 - 5.00 \times 10^{-4}v

2.78×10−4+8.70×10−5v=0.00320−5.00×10−4v2.78 \times 10^{-4} + 8.70 \times 10^{-5}v = 0.00320 - 5.00 \times 10^{-4}v

5.87×10−4v=2.92×10−35.87 \times 10^{-4}v = 2.92 \times 10^{-3}

v=4.97 cm3v = 4.97\ \text{cm}^3

Maximum volume of HCl: 4.97 cm34.97\ \text{cm}^3.