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Acids, Bases and Salts -- Diagnostic Tests

DSE Chemistry Diagnostic: Acids, Bases and Salts

Section titled “DSE Chemistry Diagnostic: Acids, Bases and Salts”

Unit Test 1: Weak Acid pH Calculation

Section titled “Unit Test 1: Weak Acid pH Calculation”

Question

Ethanoic acid (CH3COOHCH_{3}COOH) is a weak acid with Ka=1.8×10−5K_{a} = 1.8 \times 10^{-5} mol/dm3^{3} at 25∘^{\circ}C.

(a) Calculate the pH of a 0.10 mol/dm3^{3} solution of ethanoic acid. [4 marks]

(b) Calculate the pH of a 0.010 mol/dm3^{3} solution of ethanoic acid. [2 marks]

(c) A student claims that diluting a weak acid by a factor of 10 will increase the pH by exactly 1. Evaluate this claim by comparing your answers to (a) and (b). [2 marks]

Worked Solution

(a) CH3COOH⇌CH3COO−+H+CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H^{+}

Ka=[CH3COO−][H+][CH3COOH]K_{a} = \frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}

At equilibrium, let [H+]=x[H^{+}] = x:

1.8×10−5=x⋅x0.10−x1.8 \times 10^{-5} = \frac{x \cdot x}{0.10 - x}

Assuming x≪0.10x \ll 0.10 (i.e., dissociation is small):

1.8×10−5≈x20.101.8 \times 10^{-5} \approx \frac{x^{2}}{0.10} x2=1.8×10−6x^{2} = 1.8 \times 10^{-6} x=1.34×10−3 mol/dm3x = 1.34 \times 10^{-3} \text{ mol/dm}^{3}

Check: x/0.10=1.34%<5%x/0.10 = 1.34\% \lt 5\% — assumption valid.

pH=−log⁡(1.34×10−3)=2.87pH = -\log(1.34 \times 10^{-3}) = 2.87

(b) $$1.8 \times 10^{-5} = \frac{x^{2}}{0.010}

x^{2} = 1.8 \times 10^{-7}

x = 4.24 \times 10^{-4} \text{ mol/dm}^{3}

pH = -\log(4.24 \times 10^{-4}) = 3.37

(c) The pH change: $3.37 - 2.87 = 0.50$. The claim is **incorrect** for weak acids. Diluting by a factor of 10 increases the pH by **less than 1**. This is because dilution shifts the equilibrium to the right (Le Chatelier's principle), causing a greater fraction of the acid to dissociate. The $[H^{+}]$ does not decrease by a full factor of 10. (For a strong acid, the claim would be correct since $[H^{+}]$ directly decreases by a factor of 10, increasing pH by exactly 1.) --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "Diagnostics", "url": "https://dse.wyattau.com/chemistry/diagnostics"}, {"name": "Diag Acids Bases", "url": "https://dse.wyattau.com/chemistry/diagnostics/diag-acids-bases"}] } </script> ## Unit Test 2: Buffer Solution **Question** A buffer solution is prepared by mixing 100 cm$^{3}$ of 0.20 mol/dm$^{3}$ ethanoic acid ($CH_{3}COOH$, $K_{a} = 1.8 \times 10^{-5}$) with 100 cm$^{3}$ of 0.10 mol/dm$^{3}$ sodium ethanoate ($CH_{3}COONa$). (a) Calculate the pH of this buffer solution. [4 marks] (b) Calculate the new pH after adding 5.0 cm$^{3}$ of 0.10 mol/dm$^{3}$ HCl to 50.0 cm$^{3}$ of the buffer. [4 marks] (c) Explain why this buffer resists changes in pH when a small amount of strong acid is added. [2 marks] --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "Diagnostics", "url": "https://dse.wyattau.com/chemistry/diagnostics"}, {"name": "Diag Acids Bases", "url": "https://dse.wyattau.com/chemistry/diagnostics/diag-acids-bases"}] } </script> **Worked Solution** (a) After mixing, total volume = 200 cm$^{3}$.

[CH_{3}COOH] = \frac{0.20 \times 100}{200} = 0.10 \text{ mol/dm}^{3}

[CH_{3}COO^{-}] = \frac{0.10 \times 100}{200} = 0.050 \text{ mol/dm}^{3}

UsingtheHenderson−Hasselbalchequation:Using the Henderson-Hasselbalch equation:

pH = pK_{a} + \log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}

pK_{a} = -\log(1.8 \times 10^{-5}) = 4.74

pH = 4.74 + \log\frac{0.050}{0.10} = 4.74 + \log(0.5) = 4.74 - 0.30 = 4.44

(b) In 50.0 cm$^{3}$ of buffer:

n(CH_{3}COOH) = 0.10 \times 0.050 = 0.00500 \text{ mol}

n(CH_{3}COO^{-}) = 0.050 \times 0.050 = 0.00250 \text{ mol}

Moles of $HCl$ added: $n(HCl) = 0.10 \times 0.0050 = 0.000500$ mol The $H^{+}$ from HCl reacts with $CH_{3}COO^{-}$:

CH_{3}COO^{-} + H^{+} \rightarrow CH_{3}COOH}

Newmoles:New moles:

n(CH_{3}COO^{-}) = 0.00250 - 0.000500 = 0.00200 \text{ mol}

n(CH_{3}COOH) = 0.00500 + 0.000500 = 0.00550 \text{ mol}

New total volume = $50.0 + 5.0 = 55.0$ cm$^{3}$:

[CH_{3}COOH] = \frac{0.00550}{0.0550} = 0.100 \text{ mol/dm}^{3}

[CH_{3}COO^{-}] = \frac{0.00200}{0.0550} = 0.0364 \text{ mol/dm}^{3}

pH = 4.74 + \log\frac{0.0364}{0.100} = 4.74 + \log(0.364) = 4.74 - 0.439 = 4.30

(c) The buffer contains a weak acid ($CH_{3}COOH$) and its conjugate base ($CH_{3}COO^{-}$). When a strong acid ($H^{+}$) is added, the $H^{+}$ ions react with $CH_{3}COO^{-}$ to form $CH_{3}COOH$ Consuming most of the added $H^{+}$ and preventing a significant drop in pH. --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "Diagnostics", "url": "https://dse.wyattau.com/chemistry/diagnostics"}, {"name": "Diag Acids Bases", "url": "https://dse.wyattau.com/chemistry/diagnostics/diag-acids-bases"}] } </script> ## Unit Test 3: Salt Hydrolysis **Question** (a) Predict whether an aqueous solution of ammonium chloride ($NH_{4}Cl$) is acidic, alkaline, or neutral. Explain your answer using the concept of hydrolysis. [3 marks] (b) Predict whether an aqueous solution of sodium ethanoate ($CH_{3}COONa$) is acidic, alkaline, or neutral. Explain. [3 marks] (c) Predict whether an aqueous solution of sodium chloride ($NaCl$) is acidic, alkaline, or neutral. Explain. [1 mark] --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "Diagnostics", "url": "https://dse.wyattau.com/chemistry/diagnostics"}, {"name": "Diag Acids Bases", "url": "https://dse.wyattau.com/chemistry/diagnostics/diag-acids-bases"}] } </script> **Worked Solution** (a) $NH_{4}Cl$ solution is **acidic** ($pH \lt 7$). $NH_{4}Cl$ dissociates completely in water: $NH_{4}Cl \rightarrow NH_{4}^{+} + Cl^{-}$. The ammonium ion ($NH_{4}^{+}$) is the conjugate acid of the weak base ammonia ($ NH_{3}$). It undergoes **hydrolysis**:

NH_{4}^{+} + H_{2}O \rightleftharpoons NH_{3} + H_{3}O^{+}

This reaction releases $H_{3}O^{+}$ ions, making the solution acidic. The chloride ion ($Cl^{-}$) is the conjugate base of a strong acid ($HCl$) and does not hydrolyse. (b) $CH_{3}COONa$ solution is **alkaline** ($pH \gt 7$). $CH_{3}COONa$ dissociates completely: $CH_{3}COONa \rightarrow CH_{3}COO^{-} + Na^{+}$. The ethanoate ion ($CH_{3}COO^{-}$) is the conjugate base of the weak acid ethanoic acid ($CH_{3}COOH$). It undergoes hydrolysis:

CH_{3}COO^{-} + H_{2}O \rightleftharpoons CH_{3}COOH + OH^{-}

This reaction produces $OH^{-}$ ions, making the solution alkaline. The sodium ion ($Na^{+}$) does not hydrolyse. (c) $NaCl$ solution is **neutral** ($pH = 7$). Both $Na^{+}$ (conjugate acid of the strong base $NaOH$) and $Cl^{-}$ (conjugate base of the strong acid $HCl$) do not undergo hydrolysis. Neither ion affects the $pH$ of the solution. --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "Diagnostics", "url": "https://dse.wyattau.com/chemistry/diagnostics"}, {"name": "Diag Acids Bases", "url": "https://dse.wyattau.com/chemistry/diagnostics/diag-acids-bases"}] } </script> ## Intuition **The proton shuffle:** Acids donate protons (H⁺), bases accept them, it's like a game of hot potato where protons are passed between molecules. pH measures how many free protons are floating around. **Why it matters:** From stomach acid to blood buffers, acid-base chemistry keeps biological systems alive. Understanding pH helps design medicines, treat water, and control industrial processes. **The key insight:** Buffers resist pH change by absorbing or releasing protons, they're the body's way of maintaining equilibrium despite constant disturbances. ## Integration Test 1: Titration Curve + Indicator Choice **Question** 25.0 cm$^{3}$ of 0.100 mol/dm$^{3}$ ammonia solution ($NH_{3}$, $K_{b} = 1.8 \times 10^{-5}$) is titrated with 0.100 mol/dm$^{3}$ hydrochloric acid. (a) Calculate the pH of the ammonia solution before any acid is added. [3 marks] (b) Calculate the pH at the equivalence point. [3 marks] (c) State and explain the most suitable indicator for this titration. [2 marks] --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "Diagnostics", "url": "https://dse.wyattau.com/chemistry/diagnostics"}, {"name": "Diag Acids Bases", "url": "https://dse.wyattau.com/chemistry/diagnostics/diag-acids-bases"}] } </script> **Worked Solution** (a) $NH_{3} + H_{2}O \rightleftharpoons NH_{4}^{+} + OH^{-}$

K_{b} = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} = 1.8 \times 10^{-5}

Let $[OH^{-}] = x$:

1.8 \times 10^{-5} = \frac{x^{2}}{0.100}

x = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} \text{ mol/dm}^{3}

pOH = -\log(1.34 \times 10^{-3}) = 2.87

pH = 14 - 2.87 = 11.13

(b) At the equivalence point, all $NH_{3}$ has been converted to $NH_{4}^{+}$. Moles of $NH_{3}$ = $0.100 \times 0.0250 = 0.00250$ mol Volume of HCl needed = $25.0$ cm$^{3}$ (equimolar), total volume = $50.0$ cm$^{3}$.

[NH_{4}^{+}] = \frac{0.00250}{0.0500} = 0.0500 \text{ mol/dm}^{3}

$NH_{4}^{+}$ hydrolyses: $NH_{4}^{+} + H_{2}O \rightleftharpoons NH_{3} + H_{3}O^{+}$

K_{a} = \frac{K_{w}}{K_{b}} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}

5.56 \times 10^{-10} = \frac{x^{2}}{0.0500}

x = \sqrt{2.78 \times 10^{-11}} = 5.27 \times 10^{-6}

pH = -\log(5.27 \times 10^{-6}) = 5.28

(c) The equivalence point pH is 5.28, which is **acidic**. The most suitable indicator is one whose colour change range includes pH 5.28. **Methyl orange** (pH range 3.1--4.4) is too low. **Bromocresol green** (pH range 3.8--5.4) would be suitable. **Methyl red** (pH range 4.4--6.2) is also a good choice. Phenolphthalein (pH range 8.3--10.0) would NOT be suitable as the colour change occurs well above the equivalence point pH. --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "Diagnostics", "url": "https://dse.wyattau.com/chemistry/diagnostics"}, {"name": "Diag Acids Bases", "url": "https://dse.wyattau.com/chemistry/diagnostics/diag-acids-bases"}] } </script> ## Integration Test 2: pH Mixing and Neutralisation **Question** (a) Calculate the pH when 10.0 cm$^{3}$ of 0.100 mol/dm$^{3}$ NaOH is added to 40.0 cm$^{3}$ of 0.100 mol/dm$^{3}$ HCl. [3 marks] (b) Calculate the pH when 30.0 cm$^{3}$ of 0.100 mol/dm$^{3}$ NaOH is added to 40.0 cm$^{3}$ of 0.100 mol/dm$^{3}$ HCl. [3 marks] (c) Explain why the pH changes much more dramatically between the two scenarios in (a) and (b) than between adding 10.0 cm$^{3}$ and 20.0 cm$^{3}$ of NaOH. [2 marks] --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "Diagnostics", "url": "https://dse.wyattau.com/chemistry/diagnostics"}, {"name": "Diag Acids Bases", "url": "https://dse.wyattau.com/chemistry/diagnostics/diag-acids-bases"}] } </script> **Worked Solution** (a) Moles of $HCl$: $0.100 \times 0.0400 = 0.00400$ mol Moles of NaOH: $0.100 \times 0.0100 = 0.00100$ mol $HCl$ is in excess by: $0.00400 - 0.00100 = 0.00300$ mol Total volume = $40.0 + 10.0 = 50.0$ cm$^{3}$

[H^{+}] = \frac{0.00300}{0.0500} = 0.0600 \text{ mol/dm}^{3}

pH = -\log(0.0600) = 1.22

(b) Moles of $HCl$: $0.00400$ mol Moles of NaOH: $0.100 \times 0.0300 = 0.00300$ mol $HCl$ is in excess by: $0.00400 - 0.00300 = 0.00100$ mol Total volume = $40.0 + 30.0 = 70.0$ cm$^{3}$

[H^{+}] = \frac{0.00100}{0.0700} = 0.01429 \text{ mol/dm}^{3}

pH = -\log(0.01429) = 1.85

(c) The pH changes from 1.22 to 1.85 (a change of 0.63) when NaOH added increases from 10.0 to 30.0 cm$^{3}$. The change per 10 cm$^{3}$ is relatively small because the solution still contains a large excess of $H^{+}$ (a strong acid). The pH is in the region where the logarithmic scale compresses large changes in $[H^{+}]$ into small changes in pH. However, near the equivalence point (at 40.0 cm$^{3}$ of NaOH), adding even a small amount of NaOH causes a **dramatic** pH change because the $[H^{+}]$ approaches very small values where the logarithmic relationship amplifies the change. This is the characteristic **steep region** of a strong acid-strong base titration curve. --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "Diagnostics", "url": "https://dse.wyattau.com/chemistry/diagnostics"}, {"name": "Diag Acids Bases", "url": "https://dse.wyattau.com/chemistry/diagnostics/diag-acids-bases"}] } </script> ## Integration Test 3: Buffer Preparation and Capacity **Question** (a) Describe how you would prepare 250 cm$^{3}$ of an ethanoic acid / sodium ethanoate buffer with pH = 5.00, using 0.50 mol/dm$^{3}$ ethanoic acid and solid sodium ethanoate ($M = 82.0$ g/mol). ($K_{a} = 1.8 \times 10^{-5}$) [4 marks] (b) Calculate the mass of sodium ethanoate required. [2 marks] (c) Explain what is meant by **buffer capacity** and state how it can be increased. [2 marks] --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "Diagnostics", "url": "https://dse.wyattau.com/chemistry/diagnostics"}, {"name": "Diag Acids Bases", "url": "https://dse.wyattau.com/chemistry/diagnostics/diag-acids-bases"}] } </script> **Worked Solution** (a) Using the Henderson-Hasselbalch equation:

pH = pK_{a} + \log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}

5.00 = 4.74 + \log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}

\log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]} = 0.26

\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]} = 10^{0.26} = 1.82

The buffer needs $[CH_{3}COO^{-}]/[CH_{3}COOH] = 1.82$. To prepare 250 cm$^{3}$ (0.250 dm$^{3}$): choose $[CH_{3}COOH] = 0.20$ mol/dm$^{3}$ (from the 0.50 mol/dm$^{3}$ stock by dilution). Volume of stock needed: $V = \frac{0.20 \times 250}{0.50} = 100$ cm$^{3}$ of 0.50 mol/dm$^{3}$ $CH_{3}COOH$ Diluted to 250 cm$^{3}$. $[CH_{3}COO^{-}] = 1.82 \times 0.20 = 0.364$ mol/dm$^{3}$ $n(CH_{3}COO^{-}) = 0.364 \times 0.250 = 0.0910$ mol (b) $$m(CH_{3}COONa) = 0.0910 \times 82.0 = 7.46 \text{ g}

Procedure: Dissolve 7.46 g of sodium ethanoate in approximately 150 cm3^{3} of distilled water. Add 100 cm3^{3} of 0.50 mol/dm3^{3} ethanoic acid. Transfer to a 250 cm3^{3} volumetric flask and make up to the mark with distilled water. Mix thoroughly.

(c) Buffer capacity is the amount of strong acid or strong base that can be added to a buffer solution before the pH changes significantly ( defined as the amount needed to change the pH by 1 unit).

Buffer capacity can be increased by:

  1. Increasing the total concentration of the weak acid and its conjugate base (higher concentrations provide more H+H^{+} or OH−OH^{-} to absorb).
  2. Keeping the ratio [A−]/[HA][A^{-}]/[HA] close to 1 (maximum buffer capacity occurs when pH=pKapH = pK_{a}).

Common Mistakes

Section titled “Common Mistakes”

Assuming diluting a weak acid by 10x increases pH by exactly 1: This only applies to strong acids. For weak acids, dilution shifts the equilibrium right (Le Chatelier’s principle), so more acid dissociates and pH increases by less than 1.

Confusing buffer capacity with buffer pH: Buffer capacity is how much acid/base the buffer can absorb before pH changes significantly. A buffer with pH 5.00 can have different capacities depending on the total concentrations of the acid and conjugate base.

Forgetting that salt hydrolysis depends on the strength of the parent acid and base: Salts of strong acid + strong base are neutral. Salts of weak acid + strong base are alkaline. Salts of strong acid + weak base are acidic. Don’t assume all salts are neutral.

Cross-References

Section titled “Cross-References”