EduNinja
(a)

Calcium hydroxide is slightly soluble in water.

[ 4 ]
(i)

Write an equation to show the dissociation of calcium hydroxide, Ca(OH)2( s)\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{~s}), in aqueous solution. Include state symbols. ⇌

[ 1 ]
(ii)

Calculate the solubility, in moldm−3\mathrm{mol} \mathrm{dm}^{-3}, of Ca(OH)2\mathrm{Ca}(\mathrm{OH})_{2}.
[Ksp:Ca(OH)2,5.02×10−6 mol3dm−9]\left[K_{\mathrm{sp}}: \mathrm{Ca}(\mathrm{OH})_{2}, 5.02 \times 10^{-6} \mathrm{~mol}^{3} \mathrm{dm}^{-9}\right]
solubility = moldm−3\mathrm{mol} \mathrm{dm}^{-3}

[ 2 ]
(iii)

Suggest how the solubility of Ca(OH)2\mathrm{Ca}(\mathrm{OH})_{2} in aqueous NaOH compares to its solubility in water.

Explain your reasoning.

[ 1 ]
[Maximum number: 1]

Propanone, CH3COCH3\mathrm{CH}_{3} \mathrm{COCH}_{3}, reacts with iodine, I2\mathrm{I}_{2}, in the presence of an acid catalyst.

CH3COCH3+I2→CH3COCH2I+H++I−\mathrm{CH}_{3} \mathrm{COCH}_{3}+\mathrm{I}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{I}+\mathrm{H}^{+}+\mathrm{I}^{-}

The rate equation for this reaction is shown.

 rate =k[CH3COCH3][H+]\text { rate }=k\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\left[\mathrm{H}^{+}\right]
(a)

A four-step mechanism is suggested for the overall reaction.

CH3COCH3+I2→CH3COCH2I+H++I− rate =k[CH3COCH3][H+]\mathrm{CH}_{3} \mathrm{COCH}_{3}+\mathrm{I}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{I}+\mathrm{H}^{+}+\mathrm{I}^{-} \quad \text { rate }=k\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\left[\mathrm{H}^{+}\right]

Part of this mechanism is shown.

step 1: CH3COCH3+H+→CH3C+(OH)CH3\mathrm{CH}_{3} \mathrm{COCH}_{3}+\mathrm{H}^{+} \rightarrow \mathrm{CH}_{3} \mathrm{C}^{+}(\mathrm{OH}) \mathrm{CH}_{3}

step 2: CH3C+(OH)CH3→CH3C(OH)=CH2+H+\quad \mathrm{CH}_{3} \mathrm{C}^{+}(\mathrm{OH}) \mathrm{CH}_{3} \rightarrow \mathrm{CH}_{3} \mathrm{C}(\mathrm{OH})=\mathrm{CH}_{2}+\mathrm{H}^{+}

step 3: →

step 4: CH3C+(OH)CH2I→CH3COCH2I+H+\quad \mathrm{CH}_{3} \mathrm{C}^{+}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{I} \rightarrow \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{I}+\mathrm{H}^{+}

[ 1 ]
(i)

Identify one conjugate acid-conjugate base pair in the mechanism. conjugate acid conjugate base

[ 1 ]
[Maximum number: 3]

Iodine is found naturally in compounds in many different oxidation states.

(a)

An orange precipitate of HgI2\mathrm{HgI}_{2} forms when Hg2+\mathrm{Hg}^{2+} ions are added to KI(aq). The solubility of HgI2\mathrm{HgI}_{2} at 25∘C25^{\circ} \mathrm{C} is 1.00×10−7gdm−31.00 \times 10^{-7} \mathrm{gdm}^{-3}.

Calculate the solubility product, KspK_{\mathrm{sp}}, of HgI2\mathrm{HgI}_{2}. Include units in your answer.
[Mr:HgI2,454.4]\left[M_{\mathrm{r}}: \mathrm{HgI}_{2}, 454.4\right]
value of Ksp=K_{s p}=
units =

[ 3 ]
(a)

Calcium hydroxide is slightly soluble in water.

[ 4 ]
(i)

Write an equation to show the dissociation of calcium hydroxide, Ca(OH)2( s)\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{~s}), in aqueous solution. Include state symbols. ⇌

[ 1 ]
(ii)

Calculate the solubility, in moldm−3\mathrm{mol} \mathrm{dm}^{-3}, of Ca(OH)2\mathrm{Ca}(\mathrm{OH})_{2}.
[Ksp:Ca(OH)2,5.02×10−6 mol3dm−9]\left[K_{\mathrm{sp}}: \mathrm{Ca}(\mathrm{OH})_{2}, 5.02 \times 10^{-6} \mathrm{~mol}^{3} \mathrm{dm}^{-9}\right]
solubility = moldm−3\mathrm{mol} \mathrm{dm}^{-3}

[ 2 ]
(iii)

Suggest how the solubility of Ca(OH)2\mathrm{Ca}(\mathrm{OH})_{2} in aqueous NaOH compares to its solubility in water.

Explain your reasoning.

[ 1 ]
(a)

The solubility of Be(OH)2\mathrm{Be}(\mathrm{OH})_{2} in water is 2.40×10−6gdm−32.40 \times 10^{-6} \mathrm{gdm}^{-3} at 298 K .

[ 4 ]
(i)

Write an expression for the solubility product, KspK_{\mathrm{sp}}, of Be(OH)2\mathrm{Be}(\mathrm{OH})_{2} and state its units.

Ksp=K_{s p}=
units =
[ 2 ]
(ii)

Calculate the numerical value of KspK_{\mathrm{sp}} for Be(OH)2\mathrm{Be}(\mathrm{OH})_{2} at 298 K .

Ksp=K_{s p}=
[ 2 ]
(a)

The solubility product, KspK_{\mathrm{sp}}, of BaSO4\mathrm{BaSO}_{4} is 1.08×10−10 mol2dm−61.08 \times 10^{-10} \mathrm{~mol}^{2} \mathrm{dm}^{-6} at 298 K .

Calculate the solubility of BaSO4\mathrm{BaSO}_{4} in g per 100 cm3100 \mathrm{~cm}^{3} of solution.
solubility of BaSO4=\mathrm{BaSO}_{4}= g per 100 cm3100 \mathrm{~cm}^{3} of solution

[ 2 ]
[Maximum number: 1]

Hydrogen iodide, HI, is a colourless gas at room temperature.

(a)

In the laboratory, HI(aq) can be formed in a two-step process.

step 13I2( s)+2P(s)→2PI3( s)13 \mathrm{I}_{2}(\mathrm{~s})+2 \mathrm{P}(\mathrm{s}) \rightarrow 2 \mathrm{PI}_{3}(\mathrm{~s})

step 2PI3( s)+3H2O(l)→H3PO3(aq)+3HI(aq)2 \mathrm{PI}_{3}(\mathrm{~s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(\mathrm{aq})+3 \mathrm{HI}(\mathrm{aq})

[ 1 ]
(i)

Give the formula of the conjugate base of H3PO3\mathrm{H}_{3} \mathrm{PO}_{3}.

[ 1 ]
[Maximum number: 3]

Sodium oxide, Na2O\mathrm{Na}_{2} \mathrm{O}, is a white crystalline solid with a high melting point.

(a)

When sodium oxide reacts with water an alkaline solution is obtained.

[ 3 ]
(i)

Calculate the pH of the solution obtained when 3.10 g of sodium oxide are added to 400 cm3400 \mathrm{~cm}^{3} of water.

[ 3 ]
[Maximum number: 1]

Sulfides are compounds that contain sulfur but not oxygen.

(a)

Hydrogen sulfide gas, H2 S( g)\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}), is slightly soluble in water. It acts as a weak acid in aqueous solution.

[ 1 ]
(i)

Give the formula of the conjugate base of H2 S\mathrm{H}_{2} \mathrm{~S}.

[ 1 ]
(a)

The solubility of Sr(OH)2\mathrm{Sr}(\mathrm{OH})_{2} is 3.37×10−2 moldm−33.37 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3} at 0∘C0^{\circ} \mathrm{C}.

[ 3 ]
(i)

Write an expression for the solubility product of Sr(OH)2\mathrm{Sr}(\mathrm{OH})_{2}.

Ksp=K_{s p}=
[ 1 ]
(ii)

Calculate the value of KspK_{\mathrm{sp}} at 0∘C0^{\circ} \mathrm{C}. Include units in your answer.

Ksp=K_{s p}=

units =

[ 2 ]
0