Reactivity 3: What are the mechanisms of chemical change? practice questions.

A weak acid has Ka = 1.0 x 10^-5. What is its pKa?

• 5
• -5
• 9
• 1.0 x 10^-5

Answer: 5

pKa = -log(Ka) = -log(1.0 x 10^-5) = 5. A smaller Ka (larger pKa) means a weaker acid.

Using pH = pKa + log([A-]/[HA]), what is the pH of a buffer with equal concentrations of a weak acid (pKa = 4.76) and its conjugate base?

• 4.76
• 9.52
• 0
• 1.0

Answer: 4.76

When [A-] = [HA], the ratio [A-]/[HA] = 1 and log(1) = 0. So pH = pKa + 0 = 4.76. A buffer is most effective when the concentrations of acid and conjugate base are equal.

According to the Bronsted-Lowry theory, what is the conjugate base of the ammonium ion, NH4+?

• NH3
• NH2-
• N2
• NH4OH

Answer: NH3

A conjugate base is formed by losing one proton (H+). NH4+ - H+ gives NH3. NH4+ and NH3 form a conjugate acid-base pair, differing by one proton.

In a voltaic (galvanic) cell, at which electrode does oxidation occur and what is its charge?

• Cathode, positive
• Anode, negative
• Anode, positive
• Cathode, negative

Answer: Anode, negative

Oxidation always occurs at the anode. In a voltaic cell the anode is the negative electrode because oxidation releases electrons there, which flow through the external circuit to the cathode (positive). Remember: an OX (anode oxidation), red cat (reduction cathode).

Given E(Zn2+/Zn) = -0.76 V and E(Cu2+/Cu) = +0.34 V, what is the standard cell potential of a Zn/Cu cell? Use E(cell) = E(cathode) - E(anode).

• +1.10 V
• -0.42 V
• +0.42 V
• -1.10 V

Answer: +1.10 V

Copper has the higher (more positive) potential, so it is the cathode (reduction) and zinc is the anode (oxidation). E(cell) = E(cathode) - E(anode) = (+0.34) - (-0.76) = +1.10 V. A positive value means the reaction is spontaneous.

Methane reacts with chlorine in UV light. What type of mechanism is this, and what is the reactive intermediate?

• Nucleophilic substitution; a carbocation intermediate
• Free radical substitution; free radicals
• Electrophilic addition; carbocation
• Condensation; ester

Answer: Free radical substitution; free radicals

The reaction CH4 + Cl2 -> CH3Cl + HCl in UV light proceeds by free radical substitution. UV light homolytically splits Cl2 into chlorine radicals (initiation), which then react in propagation steps. The intermediates are highly reactive free radicals with unpaired electrons.