IB Chemistry HL 3.1 Proton Transfer Reactions

[Maximum number: 9]

Bleaches in which chlorine is the active ingredient are the most common, although some environmental groups have concerns about their use.

(a)

In aqueous chlorine the equilibrium below produces chloric(I) acid (hypochlorous acid), HOCl , the active bleach.

\mathrm{Cl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{HOCl}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})

[ 9 ]

(i)

Chloric(I) acid is a weak acid, but hydrochloric acid is a strong acid. Outline how this is indicated in the equation above.

[ 1 ]

(ii)

State a balanced equation for the reaction of chloric(I) acid with water.

[ 1 ]

$\quad \mathrm{HOCl}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{ClO}^{-}(\mathrm{aq}) / \mathrm{HOCl}(\mathrm{aq})+\mathrm{H}_{2}\mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(\mathrm{aq})+\mathrm{ClO}^{-}(\mathrm{aq})$ ;

(iii)

Partial neutralization of chloric(I) acid creates a buffer solution. Given that the $\mathrm{p} K_{\mathrm{a}}$ of chloric(I) acid is 7.53 , determine the pH of a solution that has $[\mathrm{HOCl}]=0.100 \mathrm{~mol} \mathrm{dm}^{-3}$ and $\left[\mathrm{ClO}^{-}\right]=0.0500 \mathrm{~mol} \mathrm{dm}{ }^{-3}$ .

[ 4 ]

$K_{\mathrm{a}}=10^{-7.53}=2.95 \times 10^{-8}\,\mathrm{mol}\,\mathrm{dm}^{-3}$ ;
$K_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{ClO}^{-}\right]}{[\mathrm{HOCl}]}=\frac{\left[\mathrm{H}^{+}\right](0.05)}{(0.1)} \approx \frac{\left[\mathrm{H}^{+}\right]}{2}=2.95 \times 10^{-8}\,\mathrm{mol}\,\mathrm{dm}^{-3}$ ;
$\left[\mathrm{H}^{+}\right]=2 \times 2.95 \times 10^{-8}=5.9 \times 10^{-8}\,\mathrm{mol}\,\mathrm{dm}^{-3}$ ;
$\mathrm{pH}=-\log\left(5.9 \times 10^{-8}\right)=7.23$ ;

(iv)

Describe, using HIn to represent the indicator in its acid form, why an indicator changes colour when excess alkali is added.

[ 3 ]

$\mathrm{HIn} \rightleftharpoons \mathrm{H}^{+}+\mathrm{In}^{-}$ ;
(weak acid in which the) acid/HIn and conjugate base/In- have different colours / OWTTE;
excess alkali shifts the equilibrium to the RHS/towards the conjugate base;

[Maximum number: 5]

Nitrous acid, $\mathrm{HNO}_{2}$ , is a weak acid which can be used to make acidic buffers.

(a)

A $1.00 \mathrm{~mol} \mathrm{dm}^{-3}$ solution of nitrous acid was prepared. This solution contains two conjugate acid-base pairs.

[ 5 ]

(i)

State the formulas of the conjugate acid and conjugate base in each pair.

Conjugate acid:
Conjugate base:
Conjugate acid:
Conjugate base:

[ 2 ]

conjugate acid $\mathrm{H}_{3} \mathrm{O}^{+}$ «（aq）» AND conjugate base $\mathrm{H}_{2} \mathrm{O}$ « $(\mathrm{l}) »$
conjugate acid $\mathrm{HNO}_{2}$ «（aq）» AND conjugate base $\mathrm{NO}_{2}{ }^{-}$ «（aq）»

(ii)

This $1.00 \mathrm{~mol} \mathrm{dm}^{-3}$ solution of nitrous acid was used to prepare a buffer with pH 3.00 .

Calculate the concentration of the conjugate base of nitrous acid required to make this buffer. The $\mathrm{pK}_{\mathrm{a}}$ of nitrous acid is 3.25.

Concentration of conjugate base:

[ 3 ]

$K_{a}<10^{-3.25}>=5.62 \times 10^{-4}$

« $\left[\mathrm{H}^{+}\right]=10^{-3.00}$ » =0.001 «mol dm ${ }^{-3}$ »
$\text { « }\left[\mathrm{A}^{-}\right]=\left(5.62 \times 10^{-4} \times 1.00\right) / 0.001 »=0.562 \text { «mol } \mathrm{dm}^{-3} \text { » }$

Alternative solution：
$\mathrm{pH}=\mathrm{p} K_{a}+\log _{10}[\text { salt }] /[\text { acid }]$ $\text { « } \log _{10}[\text { salt }] /[\text { acid }] \text { 》 }=-0.25$

OR
«[salt]/[acid]» = 0.562、

《 $\left[\mathrm{A}^{-}\right]=0.562 / 1.00 »=0.562$ «mol $\mathrm{dm}^{-3}$ » ↓

Award［3］for correct final answer．

[Maximum number: 4]

Phosphine (IUPAC name phosphane) is a hydride of phosphorus, with the formula $\mathrm{PH}_{3}$ .

(a)

Ammonia acts as a weak Brønsted-Lowry base when dissolved in water.

\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{NH}_{4}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})

Outline what is meant by the terms "weak" and "Brønsted-Lowry base".

Weak:

Brønsted-Lowry base:

[ 2 ]

(b)

Phosphine is usually prepared by heating white phosphorus, one of the allotropes of phosphorus, with concentrated aqueous sodium hydroxide. The equation for the reaction is:

\mathrm{P}_{4}(\mathrm{~s})+3 \mathrm{OH}^{-}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{PH}_{3}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{PO}_{2}^{-}(\mathrm{aq})

[ 2 ]

(i)

The ion $\mathrm{H}_{2} \mathrm{PO}_{2}^{-}$ is amphiprotic. Outline what is meant by amphiprotic, giving the formulas of both species it is converted to when it behaves in this manner.

[ 2 ]

can act as both a «Brønsted-Lowry» acid and a «Brønsted-Lowry» base OR can accept and/or donate a hydrogen ion/proton/ $\mathrm{H}^{+}$ $\mathrm{HPO}_{2}{ }^{2-}$ AND $\mathrm{H}_{3} \mathrm{PO}_{2}$

[Maximum number: 1]

Bleaches in which chlorine is the active ingredient are the most common, although some environmental groups have concerns about their use.

In aqueous chlorine the equilibrium below produces chloric(I) acid (hypochlorous acid), HOCl , the active bleach.

\mathrm{Cl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{HOCl}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})

State a balanced equation for the reaction of chloric(I) acid with water.

$\quad \mathrm{HOCl}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{ClO}^{-}(\mathrm{aq}) / \mathrm{HOCl}(\mathrm{aq})+\mathrm{H}_{2}\mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(\mathrm{aq})+\mathrm{ClO}^{-}(\mathrm{aq})$ ;

[Maximum number: 2]

Nitrous acid, $\mathrm{HNO}_{2}$ , is a weak acid which can be used to make acidic buffers.

A $1.00 \mathrm{~mol} \mathrm{dm}^{-3}$ solution of nitrous acid was prepared. This solution contains two conjugate acid-base pairs.

State the formulas of the conjugate acid and conjugate base in each pair.

Conjugate acid:
Conjugate base:
Conjugate acid:
Conjugate base:

conjugate acid $\mathrm{H}_{3} \mathrm{O}^{+}$ «（aq）» AND conjugate base $\mathrm{H}_{2} \mathrm{O}$ « $(\mathrm{l}) »$
conjugate acid $\mathrm{HNO}_{2}$ «（aq）» AND conjugate base $\mathrm{NO}_{2}{ }^{-}$ «（aq）»

[Maximum number: 2]

Phosphine (IUPAC name phosphane) is a hydride of phosphorus, with the formula $\mathrm{PH}_{3}$ .

Phosphine is usually prepared by heating white phosphorus, one of the allotropes of phosphorus, with concentrated aqueous sodium hydroxide. The equation for the reaction is:

\mathrm{P}_{4}(\mathrm{~s})+3 \mathrm{OH}^{-}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{PH}_{3}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{PO}_{2}^{-}(\mathrm{aq})

The ion $\mathrm{H}_{2} \mathrm{PO}_{2}^{-}$ is amphiprotic. Outline what is meant by amphiprotic, giving the formulas of both species it is converted to when it behaves in this manner.

can act as both a «Brønsted-Lowry» acid and a «Brønsted-Lowry» base OR can accept and/or donate a hydrogen ion/proton/ $\mathrm{H}^{+}$ $\mathrm{HPO}_{2}{ }^{2-}$ AND $\mathrm{H}_{3} \mathrm{PO}_{2}$

[Maximum number: 3]

A group of students investigated the rate of the reaction between aqueous sodium thiosulfate and hydrochloric acid according to the equation below.

\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(\mathrm{aq})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow 2 \mathrm{NaCl}(\mathrm{aq})+\mathrm{SO}_{2}(\mathrm{~g})+\mathrm{S}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(1)

The two reagents were rapidly mixed together in a beaker and placed over a mark on a piece of paper. The time taken for the precipitate of sulfur to obscure the mark when viewed through the reaction mixture was recorded.

Initially they measured out $10.0 \mathrm{~cm}^{3}$ of $0.500 \mathrm{moldm}^{-3}$ hydrochloric acid and then added $40.0 \mathrm{~cm}^{3}$ of $0.0200 \mathrm{~mol} \mathrm{dm} \mathrm{m}^{-3}$ aqueous sodium thiosulfate. The mark on the paper was obscured 47 seconds after the solutions were mixed.

(a)

The teacher asked the students to devise another technique to measure the rate of this reaction.

[ 3 ]

(i)

One group suggested recording how long it takes for the pH of the solution to change by one unit. Calculate the initial pH of the original reaction mixture.

[ 2 ]

$\left[\mathrm{H}^{+}\right]=0.5 \times \frac{10}{50}=0.1\left(\mathrm{moldm}^{-3}\right);$ $\mathrm{pH}\left(=-\log \left[\mathrm{H}^{+}\right]=-\log (0.10)\right)=1 ;$

(ii)

Deduce the percentage of hydrochloric acid that would have to be used up for the pH to change by one unit.

[ 1 ]

$\quad 90 \%$ ;

[Maximum number: 6]

Iron rusts in the presence of oxygen and water. Rusting is a redox process involving several steps that produces hydrated iron(III) oxide, $\mathrm{Fe}_{2} \mathrm{O}_{3} \bullet \mathrm{nH}_{2} \mathrm{O}$ , as the final product. The half-equations involved for the first step of rusting are given below.

Half-equation 1: $\quad \mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}$

Half-equation 2: $\quad \mathrm{O}_{2}(\mathrm{aq})+4 \mathrm{e}^{-}+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 4 \mathrm{OH}^{-}(\mathrm{aq})$

(a)

State a chemical equation for the partial dissociation of water into ions, including state symbols.

[ 1 ]

$\quad \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(aq)+\mathrm{OH}^{-}(aq) / 2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(aq)+\mathrm{OH}^{-}(aq)$ ;

(b)

The dissociation of water into ions is reversible. State the expression for the ionic product constant of water.

[ 1 ]

$K_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] / K_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]$ ;

(c)

The ionic product constant of water was measured at three different temperatures.

Table for Question (c) — IB Chemistry HL

Deduce whether the ionization of water is exothermic or endothermic, giving your reason.

[ 2 ]

(d)

Use the data in part (iii) to determine the pH of water at 373 K , correct to two decimal places.

[ 2 ]

$\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]=\left[\mathrm{H}^{+}\right]=\sqrt{5.13 \times 10^{-13}}=7.16 \times 10^{-7}\,\mathrm{mol\,dm}^{-3}$ ;
pH=6.14 / 6.15;

[Maximum number: 1]

When nitrogen gas and hydrogen gas are allowed to react in a closed container the following equilibrium is established.

\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \quad \Delta H=-92.6 \mathrm{~kJ}

Define the term weak base according to the Brønsted-Lowry theory.

[Maximum number: 1]

Methanoic acid is a monoprotic weak acid.

The concentration of methanoic acid was found by titration with a $0.200 \mathrm{~mol} \mathrm{dm}^{-3}$ standard solution of sodium hydroxide, NaOH(aq), using an indicator to determine the end point.

Write an equation for the reaction of methanoic acid with sodium hydroxide.

$\mathrm{HCOOH}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{HCOONa}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

[Maximum number: 1]

Which chart shows the neutralization curve of a strong base with a strong acid?

A

Figure for Question 11 — IB Chemistry HL — Option A

B

Figure for Question 11 — IB Chemistry HL — Option B

C

Figure for Question 11 — IB Chemistry HL — Option C

D

Figure for Question 11 — IB Chemistry HL — Option D

[Maximum number: 2]

Methanoic acid is a monoprotic weak acid.

The concentration of methanoic acid was found by titration with a $0.200 \mathrm{~mol} \mathrm{dm}^{-3}$ standard solution of sodium hydroxide, NaOH(aq), using an indicator to determine the end point.

Calculate the pH of the sodium hydroxide solution.

$[\mathrm{OH^-}]=0.200\ \mathrm{mol\ dm^{-3}}$

Alternative 1:
$\mathrm{pOH}=-\log(0.200)=0.699$ $\mathrm{pH}=14.000-0.699=13.301$

Alternative 2:
$[\mathrm{H^+}]=\frac{1.00\times10^{-14}}{0.200} =5.00\times10^{-14}\ \mathrm{mol\ dm^{-3}}$ $\mathrm{pH}=-\log(5.00\times10^{-14})=13.301$

[Maximum number: 3]

Two hydrides of nitrogen are ammonia and hydrazine, $\mathrm{N}_{2} \mathrm{H}_{4}$ . One derivative of ammonia is methanamine whose molecular structure is shown below.

Figure for Question 13 — IB Chemistry HL

Hydrazine reacts with water in a similar way to ammonia. (The association of a molecule of hydrazine with a second $\mathrm{H}^{+}$ is so small it can be neglected.)

\begin{gathered} \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{N}_{2} \mathrm{H}_{5}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \\ \mathrm{pK}_{\mathrm{b}}(\text { hydrazine })=5.77 \end{gathered}

Calculate the pH of a $0.0100 \mathrm{~mol} \mathrm{dm}^{-3}$ solution of hydrazine.

$K_b=10^{-5.77}=1.698\times10^{-6}$

or:
$K_b=\frac{[\mathrm{N}_{2}\mathrm{H}_{5}^{+}][\mathrm{OH}^{-}]}{[\mathrm{N}_{2}\mathrm{H}_{4}]}$ $[\mathrm{OH}^{-}]^2 =1.698\times10^{-6}\times0.0100 =1.698\times10^{-8}$ $[\mathrm{OH}^{-}] =\sqrt{1.698\times10^{-8}} =1.303\times10^{-4}\ \mathrm{mol\ dm^{-3}}$ $\mathrm{pOH}=3.89,\qquad \mathrm{pH}=10.11$

Accept answers in the range 10.1-10.2.

[Maximum number: 4]

When nitrogen gas and hydrogen gas are allowed to react in a closed container the following equilibrium is established.

\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \quad \Delta H=-92.6 \mathrm{~kJ}

Determine the pH of a $0.100 \mathrm{~mol} \mathrm{dm}^{-3}$ solution of ammonia, $\mathrm{NH}_{3}(\mathrm{aq})$ , using tables 2 and 15 of the data booklet.

$\quad K_{\mathrm{b}}=\frac{\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{3}\right]}=1.8 \times 10^{-5} / 10^{-4.75}$ ;
[ $\mathrm{NH}_{4}^{+}$ ] $=\left[\mathrm{OH}^{-}\right]$ and $\left[\mathrm{NH}_{3}\right] \approx 1.00 \times 10^{-1}\left(\mathrm{~mol} \mathrm{dm}^{-3}\right)$ ;
$\left[\mathrm{OH}^{-}\right]=\left(\sqrt{1.8 \times 10^{-6}}=\right) 1.3 \times 10^{-3}\left(\mathrm{~mol} \mathrm{dm}^{-3}\right) / \mathrm{pOH}=2.89$ ;
pH=(14.0-2.89=) 11.1;

Award [4] for correct final answer.

[Maximum number: 1]

Phenylethene (styrene) is produced from ethylbenzene in a gas-phase equilibrium.

Figure for Question 15 — IB Chemistry HL

Benzoic acid is a weak acid. The graph shows how the pH changes during the titration of a $10.0 \mathrm{~cm}^{3}$ aqueous solution of benzoic acid with aqueous sodium hydroxide.

Explain, with reference to acid-base equilibria, why the sodium benzoate solution formed has a pH>7.

«benzoate ion» hydrolysis/reaction with water AND forms $\mathrm{OH}^{-}$ OR
$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \quad$

[Maximum number: 3]

When nitrogen gas and hydrogen gas are allowed to react in a closed container the following equilibrium is established.

\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \quad \Delta H=-92.6 \mathrm{~kJ}

Sketch the pH titration curve obtained when $50.0 \mathrm{~cm}^{3}$ of $0.100 \mathrm{moldm}^{-3} \mathrm{NH}_{3}(\mathrm{aq})$ is added to $25.0 \mathrm{~cm}^{3}$ of $0.100 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HCl}$ (aq).

Figure for Question 16 — IB Chemistry HL

Figure for Answer to Question 16 — IB Chemistry HL

For volume =0: pH=1;
vertical jump should be positioned in volume range $24 \mathrm{~cm}^{3}$ to $26 \mathrm{~cm}^{3}$ and include pH range between 3 to 6;
For volume =50 : pH between 8 to 11;

[Maximum number: 3]

Bleaches in which chlorine is the active ingredient are the most common, although some environmental groups have concerns about their use.

In aqueous chlorine the equilibrium below produces chloric(I) acid (hypochlorous acid), HOCl , the active bleach.

\mathrm{Cl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{HOCl}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})

Describe, using HIn to represent the indicator in its acid form, why an indicator changes colour when excess alkali is added.

$\mathrm{HIn} \rightleftharpoons \mathrm{H}^{+}+\mathrm{In}^{-}$ ;
(weak acid in which the) acid/HIn and conjugate base/In- have different colours / OWTTE;
excess alkali shifts the equilibrium to the RHS/towards the conjugate base;

[Maximum number: 1]

When nitrogen gas and hydrogen gas are allowed to react in a closed container the following equilibrium is established.

\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \quad \Delta H=-92.6 \mathrm{~kJ}

Identify an indicator from table 16 of the data booklet that could be used for this titration.

[Maximum number: 1]

Phenylethene (styrene) is produced from ethylbenzene in a gas-phase equilibrium.

Figure for Question 19 — IB Chemistry HL

Describe how a buffer solution could be made from aqueous benzoic acid and aqueous sodium hydroxide.

[Maximum number: 3]

Nitrous acid, $\mathrm{HNO}_{2}$ , is a weak acid which can be used to make acidic buffers.

A $1.00 \mathrm{~mol} \mathrm{dm}^{-3}$ solution of nitrous acid was prepared. This solution contains two conjugate acid-base pairs.

This $1.00 \mathrm{~mol} \mathrm{dm}^{-3}$ solution of nitrous acid was used to prepare a buffer with pH 3.00 .

Calculate the concentration of the conjugate base of nitrous acid required to make this buffer. The $\mathrm{pK}_{\mathrm{a}}$ of nitrous acid is 3.25.

Concentration of conjugate base:

$K_{a}<10^{-3.25}>=5.62 \times 10^{-4}$

« $\left[\mathrm{H}^{+}\right]=10^{-3.00}$ » =0.001 «mol dm ${ }^{-3}$ »
$\text { « }\left[\mathrm{A}^{-}\right]=\left(5.62 \times 10^{-4} \times 1.00\right) / 0.001 »=0.562 \text { «mol } \mathrm{dm}^{-3} \text { » }$

Alternative solution：
$\mathrm{pH}=\mathrm{p} K_{a}+\log _{10}[\text { salt }] /[\text { acid }]$ $\text { « } \log _{10}[\text { salt }] /[\text { acid }] \text { 》 }=-0.25$

OR
«[salt]/[acid]» = 0.562、

《 $\left[\mathrm{A}^{-}\right]=0.562 / 1.00 »=0.562$ «mol $\mathrm{dm}^{-3}$ » ↓

Award［3］for correct final answer．

IB Chemistry HL 3.1 Proton Transfer Reactions Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (iv)

Question 2

Question (a)

Question (i)

Question (ii)

Question 3

Question (a)

Question (b)

Question (i)

Question 4

Question 5

Question 6

Question 7

Question (a)

Question (i)

Question (ii)

Question 8

Question (a)

Question (b)

Question (c)

Question (d)

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20