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A Level H1 Chemistry Practice Paper 2

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Questions

TuitionGoWhere Exam Practice (AI) - Chemistry H1 A-Level

Subject: Chemistry
Level: H1 (8873)
Paper: Practice Paper 2 (Version 2 of 5)
Duration: 1 hour 30 minutes
Total Marks: 60
Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates:

Write your Name, Class, and Date in the spaces provided.
Answer all questions.
Write your answers in the spaces provided on the question paper.
You may use a scientific calculator where appropriate.
A Data Booklet is provided for reference.

Section A: Structured Questions

Answer all questions in this section.

1 Ethanoic acid, $CH_3COOH$ , is a weak organic acid commonly found in vinegar.

(a) Define the term weak acid.
[1]

(b) Write an equation, including state symbols, to represent the dissociation of ethanoic acid in water.
[1]

...................................................................................................................................................

(c) The $K_a$ value for ethanoic acid is $1.7 \times 10^{-5} \text{ mol dm}^{-3}$ at 298 K.
Calculate the pH of a $0.10 \text{ mol dm}^{-3}$ solution of ethanoic acid.
[3]

2 A student performs a titration to determine the concentration of a solution of benzoic acid, $C_6H_5COOH$ .

25.0 cm³ of the benzoic acid solution is titrated against $0.050 \text{ mol dm}^{-3}$ sodium hydroxide ( $NaOH$ ). The endpoint is reached when 22.40 cm³ of $NaOH$ has been added.

(a) Construct the balanced chemical equation for the reaction between benzoic acid and sodium hydroxide.
[1]

...................................................................................................................................................

(b) Calculate the amount, in moles, of sodium hydroxide used in the titration.
[1]

3 Carbonic acid, $H_2CO_3$ , is formed when carbon dioxide dissolves in rainwater. It is a diprotic acid.

(a) Write the expression for the first acid dissociation constant, $K_{a1}$ , of carbonic acid.
[1]

...................................................................................................................................................

(b) Explain why the second dissociation constant, $K_{a2}$ , is significantly smaller than $K_{a1}$ .
[2]

...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................

(c) Rainwater saturated with $CO_2$ has a pH of approximately 5.6. Calculate the concentration of hydrogen ions, $[H^+]$ , in this rainwater.
[1]

4 Aluminium oxide, $Al_2O_3$ , is described as an amphoteric oxide.

(a) Define the term amphoteric.
[1]

...................................................................................................................................................

(b) Write balanced ionic equations for the reaction of solid aluminium oxide with:
(i) Dilute hydrochloric acid.
[1]

...................................................................................................................................................

(ii) Aqueous sodium hydroxide.
[1]

...................................................................................................................................................

5 Buffer solutions are essential in maintaining pH stability in biological systems, such as human blood.

(a) State the components required to form an acidic buffer solution.
[1]

...................................................................................................................................................

(b) A buffer solution is prepared by mixing 0.10 mol of ethanoic acid ( $CH_3COOH$ ) and 0.10 mol of sodium ethanoate ( $CH_3COONa$ ) in 1.0 dm³ of water.
Given $K_a$ for ethanoic acid is $1.7 \times 10^{-5} \text{ mol dm}^{-3}$ , calculate the pH of this buffer solution.
[2]

(c) Explain, with the aid of an equation, how this buffer solution resists a change in pH when a small amount of strong acid ( $H^+$ ) is added.
[2]

6 The solubility product constant, $K_{sp}$ , for magnesium hydroxide, $Mg(OH)_2$ , is $1.8 \times 10^{-11} \text{ mol}^3 \text{ dm}^{-9}$ at 298 K.

(a) Write the expression for the solubility product, $K_{sp}$ , of $Mg(OH)_2$ .
[1]

...................................................................................................................................................

(b) Calculate the molar solubility of $Mg(OH)_2$ in pure water at 298 K.
[3]

(c) Explain why the solubility of $Mg(OH)_2$ decreases when it is placed in a solution of sodium hydroxide ( $NaOH$ ).
[2]

7 Propanoic acid ( $C_2H_5COOH$ ) is a weak acid with a $pK_a$ of 4.87.

(a) Calculate the value of $K_a$ for propanoic acid.
[1]

(b) Sketch the pH curve for the titration of 25.0 cm³ of $0.10 \text{ mol dm}^{-3}$ propanoic acid with $0.10 \text{ mol dm}^{-3}$ sodium hydroxide.
Label the equivalence point and the region where the solution acts as a buffer.
[3]

8 Ammonia ( $NH_3$ ) is a weak base.

(a) Write the equation for the reaction of ammonia with water.
[1]

...................................................................................................................................................

(b) A solution of ammonia has a pH of 11.0 at 298 K.
Calculate the concentration of hydroxide ions, $[OH^-]$ , in this solution.
[1]

(c) The base dissociation constant, $K_b$ , for ammonia is $1.8 \times 10^{-5} \text{ mol dm}^{-3}$ .
Calculate the initial concentration of the ammonia solution.
[3]

Section B: Data-Based and Application Questions

Answer all questions in this section.

9 Lactic acid ( $CH_3CH(OH)COOH$ ) is produced in muscles during intense exercise and is also responsible for the sour taste in fermented milk products.

A student investigates the strength of lactic acid by measuring the pH of various concentrations. The results are shown below:

Concentration / $\text{mol dm}^{-3}$	pH
0.10	2.44
0.050	2.59
0.010	2.94

(a) Using the data for the 0.10 $\text{mol dm}^{-3}$ solution, calculate the $K_a$ of lactic acid.
[3]

(b) Explain how the data supports the conclusion that lactic acid is a weak acid rather than a strong acid.
[2]

(c) In the production of yogurt, bacteria convert lactose into lactic acid. As the concentration of lactic acid increases, the pH drops. Eventually, the bacterial enzymes stop working.
Explain why a low pH causes enzymes to lose their activity.
[2]

10 Tooth enamel is primarily composed of hydroxyapatite, $Ca_5(PO_4)_3OH$ . This compound can dissolve in acidic conditions, leading to tooth decay.

The equilibrium for the dissolution of hydroxyapatite is: $Ca_5(PO_4)_3OH(s) \rightleftharpoons 5Ca^{2+}(aq) + 3PO_4^{3-}(aq) + OH^-(aq)$

(a) Toothpaste often contains fluoride ions ( $F^-$ ). Fluoride ions can replace hydroxide ions in the lattice to form fluoroapatite, $Ca_5(PO_4)_3F$ , which is less soluble than hydroxyapatite.
Explain, in terms of equilibrium, how the presence of fluoride ions helps protect tooth enamel.
[3]

(b) Sugary drinks often have a pH of around 3.0.
Explain why frequent consumption of sugary drinks accelerates tooth decay, referring to the equilibrium above.
[2]

11 An unknown monoprotic acid, HA, has a concentration of $0.050 \text{ mol dm}^{-3}$ . The pH of this solution is measured to be 2.30.

(a) Calculate the concentration of hydrogen ions, $[H^+]$ , in the solution.
[1]

(b) Determine the percentage dissociation of the acid HA.
[2]

...................................................................................................................................................

12 The table below shows the $pK_a$ values of three different carboxylic acids.

Acid	Formula	$pK_a$
Ethanoic acid	$CH_3COOH$	4.76
Chloroethanoic acid	$ClCH_2COOH$	2.86
Dichloroethanoic acid	$Cl_2CHCOOH$	1.29

(a) Explain the trend in acid strength observed in the table.
[3]

(b) Predict the pH of a $0.10 \text{ mol dm}^{-3}$ solution of chloroethanoic acid.
[3]

13 A student wishes to prepare a buffer solution with a pH of 4.76 using ethanoic acid ( $pK_a = 4.76$ ) and sodium ethanoate.

(a) What ratio of $[CH_3COO^-]$ to $[CH_3COOH]$ is required to achieve this pH?
[1]

(b) If the student adds 0.01 mol of solid NaOH to 1.0 dm³ of this buffer solution (where initial concentrations of acid and salt are both 0.10 M), calculate the new pH.
[4]

14 Calcium hydroxide, $Ca(OH)_2$ , is sparingly soluble in water. A saturated solution of calcium hydroxide is known as limewater.

(a) Write the expression for $K_{sp}$ of calcium hydroxide.
[1]

...................................................................................................................................................

(b) The solubility of $Ca(OH)_2$ decreases as temperature increases.
Deduce whether the dissolution of $Ca(OH)_2$ is exothermic or endothermic. Explain your reasoning.
[2]

15 Indicators are weak acids or bases that change color depending on the pH of the solution. Methyl orange has a $pK_{ind}$ of 3.7.

$HIn(aq) \rightleftharpoons H^+(aq) + In^-(aq)$ (Red) $\quad \quad \quad \quad \quad \quad$ (Yellow)

(a) Derive the relationship between pH, $pK_{ind}$ , and the ratio $\frac{[In^-]}{[HIn]}$ .
[2]

(b) At what pH will the concentration of the red form ( $HIn$ ) equal the concentration of the yellow form ( $In^-$ )?
[1]

(c) Explain why methyl orange is suitable for the titration of a strong acid with a weak base, but not for a weak acid with a strong base.
[2]

Section C: Extended Response

16 The concept of pH is central to understanding aqueous chemistry.

(a) Define pH.
[1]

...................................................................................................................................................

(b) Calculate the pH of a $0.050 \text{ mol dm}^{-3}$ solution of sulfuric acid, $H_2SO_4$ , assuming complete dissociation of both protons.
[2]

(c) In reality, the second dissociation of sulfuric acid is not complete ( $K_{a2} \approx 1.0 \times 10^{-2} \text{ mol dm}^{-3}$ ).
Explain qualitatively whether the actual pH of the solution would be higher or lower than the value calculated in (b).
[2]

17 Magnesium reacts with two different acids, both of concentration $1.0 \text{ mol dm}^{-3}$ :

Acid A: Hydrochloric acid ( $HCl$ )
Acid B: Ethanoic acid ( $CH_3COOH$ )

(a) Compare the initial rate of reaction of magnesium with Acid A and Acid B. Explain your answer in terms of hydrogen ion concentration.
[3]

(b) Compare the total volume of hydrogen gas produced when excess magnesium is reacted with 50 cm³ of Acid A versus 50 cm³ of Acid B. Explain your answer.
[2]

18 The solubility of silver chloride, $AgCl$ , is affected by the presence of other ions. $K_{sp}(AgCl) = 1.8 \times 10^{-10} \text{ mol}^2 \text{ dm}^{-6}$ .

(a) Calculate the solubility of $AgCl$ in pure water.
[2]

(b) Calculate the solubility of $AgCl$ in a $0.10 \text{ mol dm}^{-3}$ solution of $NaCl$ .
[3]

19 A student is given two white solids: Sodium Chloride ( $NaCl$ ) and Ammonium Chloride ( $NH_4Cl$ ).

(a) Predict the pH of an aqueous solution of each salt (acidic, alkaline, or neutral).
[2]

(b) Explain your prediction for Ammonium Chloride using an ionic equation.
[2]

20 Titration curves provide valuable information about acid-base reactions.

(a) Sketch the titration curve for the addition of $0.10 \text{ mol dm}^{-3}$ $NaOH$ to $25.0 \text{ cm}^3$ of $0.10 \text{ mol dm}^{-3}$ $HCl$ .
Indicate the pH at the start, the equivalence point, and the final pH.
[3]

(b) Suggest a suitable indicator for this titration and state its color change.
[1]

...................................................................................................................................................

End of Paper

Answers

TuitionGoWhere Exam Practice (AI) - Chemistry H1 A-Level

Marking Scheme & Answer Key (Version 2)

Subject: Chemistry H1
Paper: Practice Paper 2

Section A: Structured Questions

1
(a) A weak acid is an acid that partially dissociates (or ionizes) in water. [1]
(b) $CH_3COOH(aq) \rightleftharpoons CH_3COO^-(aq) + H^+(aq)$

Must use reversible arrow ( $\rightleftharpoons$ ). [1]
State symbols correct. [0 if missing, but usually part of the mark for the equation in this context]
(c)

Expression: $K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}$
Assumption: $[H^+] = [CH_3COO^-]$ and $[CH_3COOH]_{eq} \approx [CH_3COOH]_{initial}$
$[H^+] = \sqrt{K_a \times [CH_3COOH]} = \sqrt{1.7 \times 10^{-5} \times 0.10}$
$[H^+] = \sqrt{1.7 \times 10^{-6}} = 1.30 \times 10^{-3} \text{ mol dm}^{-3}$
$pH = -\log(1.30 \times 10^{-3}) = 2.88$ $p H = - lo g (1.30 \times 1 0^{- 3}) = 2.88$ (or 2.89) [3]
- 1 mark for substitution into correct formula.
- 1 mark for correct $[H^+]$ .
- 1 mark for correct pH.

2
(a) $C_6H_5COOH(aq) + NaOH(aq) \rightarrow C_6H_5COONa(aq) + H_2O(l)$ [1]
(b)

Volume in $dm^3 = 22.40 / 1000 = 0.0224 \text{ dm}^3$
$n(NaOH) = c \times V = 0.050 \times 0.0224 = 1.12 \times 10^{-3} \text{ mol}$ [1]
(c)
Ratio is 1:1, so $n(\text{acid}) = 1.12 \times 10^{-3} \text{ mol}$
Volume of acid = $25.0 \text{ cm}^3 = 0.025 \text{ dm}^3$
$c(\text{acid}) = \frac{1.12 \times 10^{-3}}{0.025} = 0.0448 \text{ mol dm}^{-3}$ $c (acid) = \frac{1.12 \times 1 0 ^{- 3}}{0.025} = 0.0448 mol dm^{- 3}$ [2]
- 1 mark for moles of acid.
- 1 mark for concentration.

3
(a) $K_{a1} = \frac{[HCO_3^-][H^+]}{[H_2CO_3]}$ [1]
(b)

It is harder to remove a positive proton ( $H^+$ ) from a negatively charged ion ( $HCO_3^-$ ) than from a neutral molecule ( $H_2CO_3$ ).
Electrostatic attraction between $H^+$ and $HCO_3^-$ is stronger. [2]
(c) $[H^+] = 10^{-pH} = 10^{-5.6} = 2.5 \times 10^{-6} \text{ mol dm}^{-3}$ [1]

4
(a) An amphoteric substance can act as both an acid and a base (reacts with both acids and bases). [1]
(b)
(i) $Al_2O_3(s) + 6H^+(aq) \rightarrow 2Al^{3+}(aq) + 3H_2O(l)$ [1]
(ii) $Al_2O_3(s) + 2OH^-(aq) + 3H_2O(l) \rightarrow 2[Al(OH)_4]^-(aq)$ [1]
* Accept $Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O$ if balanced correctly, but complex ion form is preferred in modern syllabi.

5
(a) A weak acid and its salt (conjugate base). [1]
(b)

$pH = pK_a + \log \left( \frac{[\text{salt}]}{[\text{acid}]} \right)$
Since $[\text{salt}] = [\text{acid}] = 0.10$ , the log term is $\log(1) = 0$ .
$pH = pK_a = -\log(1.7 \times 10^{-5}) = 4.77$ [2]
(c)
Added $H^+$ ions react with the conjugate base ( $CH_3COO^-$ ).
Equation: $CH_3COO^-(aq) + H^+(aq) \rightarrow CH_3COOH(aq)$
This removes most of the added $H^+$ , keeping pH relatively constant. [2]

6
(a) $K_{sp} = [Mg^{2+}][OH^-]^2$ [1]
(b)

Let solubility be $s \text{ mol dm}^{-3}$ .
$[Mg^{2+}] = s$ , $[OH^-] = 2s$ .
$K_{sp} = (s)(2s)^2 = 4s^3$ .
$1.8 \times 10^{-11} = 4s^3$ .
$s^3 = 4.5 \times 10^{-12}$ .
$s = \sqrt[3]{4.5 \times 10^{-12}} = 1.65 \times 10^{-4} \text{ mol dm}^{-3}$ [3]
(c)
$NaOH$ provides a high concentration of $OH^-$ ions (Common Ion Effect).
According to Le Chatelier’s Principle / $K_{sp}$ expression, increasing $[OH^-]$ shifts equilibrium to the left (precipitation), decreasing solubility. [2]

7
(a) $K_a = 10^{-pK_a} = 10^{-4.87} = 1.35 \times 10^{-5} \text{ mol dm}^{-3}$ [1]
(b)

Start pH: Weak acid, so pH > 1 (approx 2.9).
Buffer region: Gradual rise.
Equivalence point: pH > 7 (approx 8-9) because salt is basic.
Final pH: Approaches pH of NaOH (approx 13).
Shape: Sigmoidal. [3]
- 1 mark for start pH > 1.
- 1 mark for equivalence point pH > 7.
- 1 mark for general shape.

8
(a) $NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$ [1]
(b) $pOH = 14 - 11.0 = 3.0$ .
$[OH^-] = 10^{-3.0} = 1.0 \times 10^{-3} \text{ mol dm}^{-3}$ [1]
(c)

$K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}$
Assume $[NH_4^+] = [OH^-] = 1.0 \times 10^{-3}$ .
$1.8 \times 10^{-5} = \frac{(1.0 \times 10^{-3})^2}{[NH_3]_{eq}}$
$[NH_3]_{eq} = \frac{1.0 \times 10^{-6}}{1.8 \times 10^{-5}} = 0.0556 \text{ mol dm}^{-3}$
Initial $[NH_3] \approx 0.0556 \text{ mol dm}^{-3}$ (since dissociation is small). [3]

Section B: Data-Based and Application Questions

9
(a)

$pH = 2.44 \Rightarrow [H^+] = 10^{-2.44} = 3.63 \times 10^{-3} \text{ mol dm}^{-3}$
$K_a = \frac{[H^+]^2}{[HA]} = \frac{(3.63 \times 10^{-3})^2}{0.10}$
$K_a = \frac{1.32 \times 10^{-5}}{0.10} = 1.32 \times 10^{-4} \text{ mol dm}^{-3}$ [3]
(b)
If it were strong, $[H^+]$ would equal $0.10 \text{ mol dm}^{-3}$ (pH 1.0).
The actual pH is 2.44 (much higher), indicating partial dissociation. [2]
(c)
Enzymes are proteins with specific 3D structures (tertiary structure) maintained by hydrogen/ionic bonds.
Excess $H^+$ disrupts these bonds, causing denaturation. The active site changes shape, and the substrate no longer fits. [2]

10
(a)

Adding $F^-$ shifts the equilibrium: $Ca_5(PO_4)_3OH(s) + F^-(aq) \rightleftharpoons Ca_5(PO_4)_3F(s) + OH^-(aq)$ .
Since fluoroapatite is less soluble (smaller $K_{sp}$ ), the equilibrium lies far to the right, forming a more resistant layer on the tooth. [3]
(b)
$H^+$ from the acid reacts with $OH^-$ and $PO_4^{3-}$ ions.
This decreases the concentration of products, shifting the dissolution equilibrium to the right (Le Chatelier), causing more enamel to dissolve. [2]

11
(a) $[H^+] = 10^{-2.30} = 5.01 \times 10^{-3} \text{ mol dm}^{-3}$ [1]
(b)

$\% \text{ dissociation} = \frac{[H^+]_{eq}}{[HA]_{initial}} \times 100$
$= \frac{5.01 \times 10^{-3}}{0.050} \times 100 = 10.0\%$ [2]
(c) Weak acid. Strong acids dissociate ~100%. 10% is significantly less than 100%. [1]

12
(a)

Chlorine is electronegative and exerts a negative inductive effect (-I effect).
This withdraws electron density from the O-H bond, making the H more positive and easier to lose.
It also stabilizes the resulting carboxylate anion by dispersing the negative charge. More Cl atoms = stronger effect = lower $pK_a$ . [3]
(b)
$K_a = 10^{-2.86} = 1.38 \times 10^{-3}$ .
$[H^+] = \sqrt{K_a \times c} = \sqrt{1.38 \times 10^{-3} \times 0.10} = \sqrt{1.38 \times 10^{-4}} = 0.0117$ .
$pH = -\log(0.0117) = 1.93$ [3]

13
(a) When $pH = pK_a$ , the ratio $\frac{[\text{salt}]}{[\text{acid}]} = 1$ . [1]
(b)

Initial moles: Acid = 0.10, Salt = 0.10.
Add 0.01 mol NaOH. NaOH reacts with Acid.
New moles: Acid = $0.10 - 0.01 = 0.09$ . Salt = $0.10 + 0.01 = 0.11$ .
$pH = 4.76 + \log(\frac{0.11}{0.09})$ .
$pH = 4.76 + \log(1.22) = 4.76 + 0.087 = 4.85$ [4]

14
(a) $K_{sp} = [Ca^{2+}][OH^-]^2$ [1]
(b)

Solubility decreases as T increases.
This implies the reverse reaction (precipitation) is favored by heat, or the forward reaction (dissolution) is favored by cold.
Therefore, dissolution is exothermic ( $\Delta H < 0$ ). [2]

15
(a)

$K_{ind} = \frac{[H^+][In^-]}{[HIn]}$
$[H^+] = K_{ind} \frac{[HIn]}{[In^-]}$
$-\log[H^+] = -\log K_{ind} - \log(\frac{[HIn]}{[In^-]})$
$pH = pK_{ind} + \log(\frac{[In^-]}{[HIn]})$ [2]
(b) When $[HIn] = [In^-]$ , $\log(1) = 0$ , so $pH = pK_{ind} = 3.7$ . [1]
(c)
Strong Acid + Weak Base titration has an equivalence point in the acidic range (pH < 7).
Methyl orange changes color in the acidic range (3.1–4.4), matching the steep part of the curve.
Weak Acid + Strong Base has equivalence point in basic range (pH > 7), where methyl orange has already changed color. [2]

Section C: Extended Response

16
(a) $pH = -\log_{10}[H^+]$ [1]
(b)

$H_2SO_4$ is diprotic. $[H^+] = 2 \times 0.050 = 0.10 \text{ mol dm}^{-3}$ .
$pH = -\log(0.10) = 1.0$ [2]
(c)
The second dissociation is incomplete, so fewer $H^+$ ions are produced than assumed in (b).
Lower $[H^+]$ means higher pH. [2]

17
(a)

Rate is faster with HCl (Acid A).
HCl is a strong acid (fully dissociated), so $[H^+]$ is much higher ( $1.0 \text{ M}$ ) compared to ethanoic acid (weak, partial dissociation, low $[H^+]$ ).
Collision frequency between $H^+$ and Mg is higher. [3]
(b)
The total volume of gas produced is the same.
Both acids have the same volume and concentration, so they contain the same total number of moles of potential $H^+$ (stoichiometrically). Excess Mg ensures all acid reacts. [2]

18
(a)

$K_{sp} = s^2$ .
$s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5} \text{ mol dm}^{-3}$ [2]
(b)
In 0.10 M NaCl, $[Cl^-] = 0.10 \text{ M}$ .
$K_{sp} = [Ag^+][Cl^-] \Rightarrow 1.8 \times 10^{-10} = [Ag^+](0.10)$ .
$[Ag^+] = 1.8 \times 10^{-9} \text{ mol dm}^{-3}$ .
Solubility is $1.8 \times 10^{-9} \text{ mol dm}^{-3}$ . [3]
(c)
High $[Cl^-]$ from NaCl shifts equilibrium $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$ to the left.
This reduces the concentration of dissolved $Ag^+$ , lowering solubility. [2]

19
(a)

NaCl: Neutral (pH 7).
$NH_4Cl$ : Acidic (pH < 7). [2]
(b)

$NH_4^+$ is the conjugate acid of a weak base ( $NH_3$ ).
It hydrolyzes in water: $NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq)$ .
Production of $H_3O^+$ makes the solution acidic. [2]

20
(a)

Start pH: 1.0 ( $0.1 \text{ M } HCl$ ).
Equivalence point: pH 7.0 (vertical section).
End pH: ~13.0 ( $0.1 \text{ M } NaOH$ ).
Curve starts low, stays low, rises sharply at 25 cm³, levels off high. [3]
(b) Phenolphthalein (colorless to pink) or Methyl Orange (red to yellow). Both work for Strong/Strong. [1]
(c) The pH change at the equivalence point is very gradual (no steep vertical section), making it hard for an indicator to show a sharp color change. [1]

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