Acids, Bases and Salts Class 10 ICSE Important Questions with Answers

Answer: (b) $H^{+}$H^+ ions as the only positive ions.

An acid dissociates in water to give hydrogen ions (existing as hydronium ions $H_3O^{+}$H₃O^+) as the only positively charged ions, which is the source of its acidic behaviour.

Answer: (b) 7.

In a neutral solution the concentration of $H^{+}$H^+ ions equals that of $OH^{-}$OH^- ions, giving a pH of exactly 7 at room temperature.

Answer: (b) Acid salt.

It is formed by the partial replacement of the replaceable hydrogen of sulphuric acid ($H_2SO_4$H₂SO₄); one replaceable H atom still remains, so it is an acid salt.

Answer: (d) pH 12.

The higher the pH above 7, the more alkaline the solution and the greater its $OH^{-}$OH^- ion concentration. Of the values given, pH 12 is the most alkaline, so it has the highest $OH^{-}$OH^- concentration.

Answer: (a) Both statements are true and R correctly explains A. $HCl$HCl ionises almost completely to give a high $H^{+}$H^+ concentration (strong), while $CH_3COOH$CH₃COOH ionises only slightly (weak); strength depends on the degree of ionisation.

Strong acid: an acid that ionises almost completely in water to give a high concentration of $H^{+}$H^+ ions; strength refers to the degree of ionisation (e.g. $HCl$HCl, $H_2SO_4$H₂SO₄, $HNO_3$HNO₃).

Concentrated acid: an acid containing a very small amount of water (a large amount of acid per unit volume); concentration refers to the amount of water present, not to ionisation. An acid can be strong yet dilute, or weak yet concentrated.

Neutralisation is the reaction between an acid and a base to form salt and water only, in which the $H^{+}$H^+ ions of the acid combine with the $OH^{-}$OH^- ions of the base to form water.

Ionic equation:

$H^{+} + OH^{-} \rightarrow H_2O$H^+ + OH^- → H₂O

Example: $HCl + NaOH \rightarrow NaCl + H_2O$HCl + NaOH → NaCl + H₂O.

A salt is an ionic compound formed when the replaceable hydrogen of an acid is wholly or partially replaced by a metal or an ammonium ($NH_4^{+}$NH₄^+) radical.

Three types:

1. Normal salt: all the replaceable hydrogen of the acid is replaced, e.g. $NaCl$NaCl (sodium chloride).

2. Acid salt: only part of the replaceable hydrogen is replaced, so some H remains, e.g. $NaHCO_3$NaHCO₃ (sodium hydrogen carbonate).

3. Basic salt: contains hydroxide or oxide in addition to the normal salt, e.g. basic lead nitrate, $Pb(OH)NO_3$Pb(OH)NO₃.

An insoluble salt is prepared by precipitation (double decomposition): mix solutions of two suitable soluble salts, one supplying the cation and the other the anion.

Method: Add lead nitrate solution to dilute sulphuric acid (or sodium sulphate solution). A white precipitate of lead sulphate forms at once. Filter to collect the precipitate, wash it with distilled water to remove soluble impurities, and dry it between filter papers.

Equation:

$Pb(NO_3)_2 + H_2SO_4 \rightarrow PbSO_4\downarrow + 2HNO_3$Pb(NO₃)₂ + H₂SO₄ → PbSO₄↓ + 2HNO₃

Copper sulphate is a salt of the insoluble base copper(II) oxide, so it is prepared by the reaction of the insoluble base (or carbonate) with a dilute acid.

Steps:

1. Warm dilute sulphuric acid and add copper(II) oxide (black) in small amounts, stirring, until a little remains undissolved (this ensures all the acid is used up).

2. Filter to remove the excess undissolved oxide; the blue filtrate is copper sulphate solution.

3. Evaporate the filtrate to the point of crystallisation and allow it to cool slowly; blue crystals of $CuSO_4\cdot 5H_2O$CuSO₄· 5H₂O separate. Filter, wash and dry them between filter papers.

Equation:

$CuO + H_2SO_4 \rightarrow CuSO_4 + H_2O$CuO + H₂SO₄ → CuSO₄ + H₂O

(a)

(i) $Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow$Na₂CO₃+2HCl→2NaCl+H₂O+CO₂↑

(ii) $NaHCO_3+HCl\rightarrow NaCl+H_2O+CO_2\uparrow$NaHCO₃+HCl→ NaCl+H₂O+CO₂↑

(iii) $Na_2SO_3+2HCl\rightarrow2NaCl+H_2O+SO_2\uparrow$Na₂SO₃+2HCl→2NaCl+H₂O+SO₂↑ (on warming)

(iv) $NaHSO_3+HCl\rightarrow NaCl+H_2O+SO_2\uparrow$NaHSO₃+HCl→ NaCl+H₂O+SO₂↑ (on warming)

(v) $FeS+2HCl\rightarrow FeCl_2+H_2S\uparrow$FeS+2HCl→ FeCl₂+H₂S↑

(b) In (iii) the gas is sulphur dioxide, $SO_2$SO₂. It has a suffocating smell of burning sulphur and turns orange acidified potassium dichromate paper green.

In (v) the gas is hydrogen sulphide, $H_2S$H₂S. It smells of rotten eggs and turns moist lead acetate paper silvery black (lead sulphide forms).

(Carbonates and hydrogen carbonates give $CO_2$CO₂, which turns lime water milky.)

(i) $NaCl$NaCl — Method C (neutralisation/titration): sodium chloride is a soluble salt of a strong acid and strong alkali.
$NaOH + HCl \rightarrow NaCl + H_2O$NaOH + HCl → NaCl + H₂O

(ii) $FeSO_4$FeSO₄ — Method B (displacement of hydrogen from acid by an active metal):
$Fe + H_2SO_4 \rightarrow FeSO_4 + H_2\uparrow$Fe + H₂SO₄ → FeSO₄ + H₂↑

(iii) $FeCl_3$FeCl₃ — Method A (direct combination of the elements): iron combines directly with chlorine (an active non-metal) to give the higher chloride.
$2Fe + 3Cl_2 \rightarrow 2FeCl_3$2Fe + 3Cl₂ → 2FeCl₃

(iv) $PbCl_2$PbCl₂ — Method E (precipitation/double decomposition): lead chloride is insoluble.
$Pb(NO_3)_2 + 2HCl \rightarrow PbCl_2\downarrow + 2HNO_3$Pb(NO₃)₂ + 2HCl → PbCl₂↓ + 2HNO₃

(v) $ZnCO_3$ZnCO₃ — Method E (precipitation/double decomposition): zinc carbonate is insoluble.
$ZnSO_4 + Na_2CO_3 \rightarrow ZnCO_3\downarrow + Na_2SO_4$ZnSO₄ + Na₂CO₃ → ZnCO₃↓ + Na₂SO₄

(i) Lemon juice (pH 2) is the most acidic. Universal indicator turns red in a strongly acidic solution.

(ii) Baking soda solution (pH 9) and sodium hydroxide solution (pH 13) are alkaline, as both have pH greater than 7.

(iii) Higher pH means lower $H^{+}$H^+ concentration, so increasing $H^{+}$H^+ concentration is: $NaOH\,(13) <$NaOH (13) < baking soda$\,(9) <$(9) < water$\,(7) <$(7) < lemon juice$\,(2)$(2).

(iv) The pH of the sodium hydroxide solution decreases (moves towards 7). The $H^{+}$H^+ ions from the acidic lemon juice neutralise the $OH^{-}$OH^- ions of the alkali ($H^{+} + OH^{-} \rightarrow H_2O$H^+ + OH^- → H₂O), reducing the alkalinity and lowering the pH.