Acids & Bases
Titrations
Proteins/Amino Acids
Galvanic/Electrolytic
Equilibrium
100
The conjugate base of the acid HPO32– is?
PO3–
100
What is the pH at the equivalence point for the titration of a weak base with a strong acid?
pH less than 7
100
How are amino acids joined?
Peptide bond
100
Which one of the following statements BEST describes the function of an anode in an electrolytic cell?
(a) The anode is the electrode at which reduction occurs.
(b) The anode is the only electrode at which OH¯ (aq) ions are produced.
(c) The anode is the electrode which attracts positive ions.
(d) The anode is the electrode that is oxidised.
(d) The anode is the electrode that is oxidised.
100
Consider the following equilibrium system, which is formed when solid white phosphorus reacts with fluorine gas to form phosphorus trifluoride vapour.
P4(s) + 6 F2(g) ⇌ 4 PF3(g)
What is the correct equilibrium constant (K) expression for the following reaction?
Ans: K=[PF3]4/[F2]6
200
Identify a conjugate acid-base pair in this reaction, and explain why it is classified as a Brønsted – Lowry acid-base reaction.
CO2 (g) + CO32– (aq) + H2O (l) ⇌
2 HCO3– (aq)
A Brønsted–Lowry acid is a proton (H⁺) donor, and a Brønsted–Lowry base is a proton acceptor.
In this reaction:
The carbonate ion (CO₃²⁻) accepts a proton (H⁺) to become hydrogen carbonate (HCO₃⁻).
Therefore, CO₃²⁻ / HCO₃⁻ is a conjugate acid–base pair.
200
What is the average amount of titre:
Sample 1: 24.65
Sample 2: 24.86
Sample 3: 25.05
Sample 4: 24.92
29.94
200
How are the secondary structures of proteins are stabilised by this intermolecular force?
Hydrogen Bonds
200
Which of the following statements regarding fuel cells is correct?
(a) Fuel cells are a type of galvanic cell.
(b) Fuel cells are a type of electrolytic cell.
(c) Fuel cells are a type of primary cell.
(d) Fuel cells are a type of secondary cell.
(a) Fuel cells are a type of galvanic cell.
200
For the reaction:
A2(g) + B(g) ⇌ 2C(g) it is found that by adding 1.5 moles of C to a 1.0 L container, an equilibrium is established in which 0.30 moles of B are found.
Calculate the value for the equilibrium constant at the temperature at the experiment was done.
Keq = 9
300
HLit(aq) + H2O(l) ⇌ Lit –(aq) + H3O+(aq)
red blue
Explain how litmus indicator works. Include details of the colour change observed in acidic and basic solution.
Litmus works because the acidic / protonated form is a different colour (red) than the basic / deprotonated form (blue).
In acidic solution the reverse reaction is favoured due to presence of protons/hydrogen ions/hydronium ions and protonated ie. red form of litmus dominates. In basic solution the equilibrium shifts to the right due to the presence of hydroxide ions and the blue form dominates
300
What is the pH solution given that [H+] = 0.1967
ph = - log10(0.1967)=0.7066
300
What is a coiled peptide chain held in place by hydrogen bonding between peptide bonds in the same chain?
α-helix
300
Calculate the standard cell potential of a voltaic cell that uses the Ag/Ag+ and Sn/Sn2+ half-cell reactions.
Identify the anode and the cathode.
Oxidation (anode): Sn(𝑠)→Sn2+(𝑎𝑞)+2e−
Reduction (cathode): Ag+(𝑎𝑞)+e−→Ag(𝑠)
300
In the following equilibrium system, nitrogen, oxygen and chlorine gases combine to produce nitrosyl chloride vapour. This equilibrium system can form at temperatures of around 400 °C.
N2(g) + O2(g) + Cl2(g) ⇌ 2 NOCl(g)
colourless colourless greenish-yellow yellow
If an equal number of moles of N2(g), O2(g) and Cl2(g) were injected into a sealed flask at 400 °C;
Explain what would happen to the forward and reverse reaction rates as the system moved to establish equilibrium.
- the forward reaction rate would initially be high, as the molecules collide to produce NOCl, and would slowly decrease as the reactants form products
- the reverse reaction rate would initially be zero, but as NOCl is produced it would begin to increase
- at equilibrium the rate of forward and reverse reactions becomes equal
400
Hydrocyanic acid, HCN(aq), is an extremely poisonous acid with a Ka value of 6.17 x 10-10. It is made by dissolving liquid or gaseous hydrogen cyanide in water. Write an equation for the ionisation of HCN in water, using the Bronsted-Lowry theory.
Bronsted-Lowry
HCN(aq) + H2O(l) ⇌ H3O+(aq) + CN-(aq)
A B CA CB
400
Potassium hydrogen iodate, KH(IO3)2, is a common primary standard used in acid-base titrations.
A sample of KH(IO3)2(s) weighing 1.218 g was dissolved in distilled water and made up to 250.0 mL in a volumetric flask. 20.00 mL aliquots of this primary standard were taken and titrated against a sodium hydroxide solution, NaOH(aq), of unknown concentration. An average titre of 23.74 mL was required to reach the end point.
The equation for the titration reaction is;
KH(IO3)2(aq) + NaOH(aq) ⇌ NaIO3(aq) + KIO3(aq) + H2O(l)
Calculate the concentration of the NaOH(aq) solution.
0.01053M
WORKING
Given reaction (1:1):
1) Molar mass of KH(IO3)2
K (39.10) + H (1.008) + 2×[I (126.90) + 3×O (16.00)] = 389.91 g/mol
2) Moles of primary standard in 250.0 mL: (n=m/MM)
n=1.218g/389.91g/mol =3.1238×10−3 mol
3) Concentration of the standard solution: (C=n/V)
C = 3.1238×10−3/0.2500 =0.012495 mol L−1
4) Moles in a 20.00 mL aliquot: (n = CV)
N aliquot = 0.012495×0.02000=2.4990×10−4 mol
5) NaOH stoichiometry (1:1) ⇒ moles NaOH used = moles in aliquot.
Titre volume VNaOH=23.74 mL=0.02374 L
6) Concentration of NaOH: (C = n/V)
CNaOH = 2.4990×10−4/ 0.02374 = 0.01053 mol L−1
400
Sketch the polypeptide for:
Ser-Tyr-Gln
N-terminus C-terminus
H2N–CH–C(=O)–NH–CH–C(=O)–NH–CH–C(=O)–OH
| | |
CH2OH –CH2–⟐–OH –CH2–CH2–CONH2
Ser Tyr Gln
400
Draw and label the parts of an operating electrolytic cell during the electrolysis of molten potassium chloride KCl (l).
400
Aluminium salts are acidic due to the presence of the hexaaqualuminate ion, [Al(H2O)6]3+ which is formed when a soluble aluminium salt is dissolved in water. This ion undergoes hydrolysis as follows:
[Al(H2O)6]3+ (aq) + H2O (l) ⇌ [Al(OH)(H2O)5]2+ (aq) + H3O+ (aq)
A solution of aluminium nitrate has a pH of 5.6.
It was found that when the aluminium nitrate solution was warmed, the pH of the solution decreased. From this information, deduce whether the forward reaction in the above equilibrium is endothermic or exothermic. Explain your reasoning:
As the pH has decreased due to an increase in the [H+], caused by an increase in temp; (1) clearly the Forward reaction has been favoured by this imposed change, (ie. higher temp).
In order for the reaction to respond in this way, (ie. shifting the equilibrium to the right),
the Forward reaction must be ENDOTHERMIC
500
Hydrocyanic acid, HCN(aq), is an extremely poisonous acid with a Ka value of 6.17 x 10^-10. It is made by dissolving liquid or gaseous hydrogen cyanide in water. What information does the value of Ka give us about hydrocyanic acid, HCN(aq)? Explain your answer.
The acid dissociation constant (Ka) gives us a quantitative measure of how much an acid ionizes (dissociates) in water.
For hydrocyanic acid (HCN):
Interpretation of the Ka value
The very small Ka value (much less than 1) means that HCN only slightly ionizes in water.
At equilibrium, most of the HCN molecules remain undissociated, and only a very small fraction forms H⁺ and CN⁻ ions.
Therefore, HCN is a weak acid.
Explanation
A strong acid has a large Ka (≫ 1), meaning it almost completely donates its protons (H⁺) to water. A weak acid has a small Ka, meaning it only partially donates protons, establishing an equilibrium that favours the reactants (undissociated acid).
Final Answer:
The Ka value of 6.17 × 10⁻¹⁰ indicates that hydrocyanic acid is a very weak acid, because it ionizes only slightly in water — only a tiny fraction of HCN molecules release protons (H⁺).
500
Lulu Labwrecker carefully pipets 25.0 mL of 0.525 M NaOH into a test tube. She places the test tube into a small beaker to keep it from spilling and then pipets 75.0 mL of 0.355 M HCl into another test tube. When Lulu reaches to put this test tube of acid into the beaker along with test tube of base she accidentally knocks the test tubes together hard enough to break them and their respective contents combine in the bottom of the beaker.
What is the pH of the resulting solution?
Is the solution formed from the contents of the two test tubes acidic or basic?
First neutralize the strong acid/base:
- Moles OH⁻ from NaOH: (n=CV)
0.0250 L×0.525 M=0.013125 mol - Moles H⁺ from HCl: (n=CV)
0.0750 L×0.355 M=0.026625 mol - Excess acid after reaction:
- n(H⁺)=n(HCl)−n(NaOH)
0.026625−0.013125=0.013500 mol H+
Total volume: 25.0 + 75.0=100.0 mL =0.1000 L
Concentration of H+ (C =n/V)
- [H+] = 0.0135000/ 0.1000=0.135 M
pH=−log10(0.135) ≈0.870
Answer:
- pH ≈ 0.870
- The resulting solution is acidic (excess HCl).
500
Identify each of the amino acids in the polypeptide and then name it using the three character abbreviations.
N-terminus C-terminus
H2N–CH–C(=O)–NH–CH–C(=O)–NH–CH–C(=O)–OH
| | |
(CH2)4–NH2 CH2–CH2–S–CH3 H
Lys-Met-Gly
500
A 1.0 M HCl solution is electrolyzed using a copper anode and an inert carbon cathode. Predict the half-reactions that will occur and describe what you would observe at each electrode.
CATHODE: 2 H+ + e- → H2 (g)
ANODE: Cu (s) → Cu2+ + 2 e-
You would observe bubbles around the cathode since H2 (g) is being produced, but the electrode would not change in mass. You would observe the anode slowly losing mass, since the Cu (s) is being oxidized into Cu2+.
You would also observe the solution turning blue since Cu2+ is being produced.
THINKING:
At the anode (oxidation)
Possible oxidations:
Chloride ions oxidized:
Copper metal oxidized:
Which occurs?
The standard electrode potential for Cu²⁺/Cu is +0.34 V, meaning copper oxidizes fairly easily.
The E° for Cl₂/Cl⁻ is +1.36 V, so chloride oxidation requires more energy.
Therefore, copper metal is preferentially oxidized, not chloride ions.
So, at the anode, copper dissolves forming Cu²⁺ ions.
At the cathode (reduction)
Possible reductions:
H⁺ ions reduced:
Cu²⁺ reduced: (but mainly at the anode we’re producing Cu²⁺, not enough yet to deposit)
Since the cathode is inert and the solution is strongly acidic,
H⁺ ions are reduced to H₂ gas.
500
Nitric oxide is an unstable molecule, and when it comes into contact with ozone the following reaction takes place, producing nitrogen dioxide (NO2) and oxygen (O2).
O3(g) + NO(g) ⇌ NO2(g) + O2(g)
The activation energy for the forward reaction is 10.8 kJ, whilst the activation energy for the reverse reaction is 210.5 kJ.
Considering the activation energy values provided;
Comment on the likely reversibility of this reaction and explain how this would affect the size of the equilibrium constant (Kc):
- not likely to be reversible, since forward Ea very small and reverse Ea very large
- this suggests there would be lots of products present at equilibrium
- this would result in a large value of Kc