Volumetric Analysis

i) Relation between normality and molarity

Let $w$ g of solute (molar mass $M_w$, equivalent weight $E$) be dissolved in $V$ litres of solution.

$$\text{Molarity } (M) = \frac{w}{M_w \times V} \qquad \text{Normality } (N) = \frac{w}{E \times V}$$

Equivalent weight = $\dfrac{\text{molar mass}}{n}$, where $n$ = n-factor (basicity of acid, acidity of base, or total valency/electrons transferred). So $E = M_w/n$:

$$N = \frac{w}{(M_w/n) \times V} = n \times \frac{w}{M_w \times V}$$ $$\therefore\; N = n \times M \quad \text{(Normality = n-factor × Molarity)}$$

Example: 1 M $H_2SO_4$ = 2 N.

ii) Resulting normality of the mixture

Given: NaOH: 20 mL of 0.8 M (n = 1), so $N = 0.8$ N. $H_2SO_4$: 25 mL of 0.4 M (n = 2), so $N = 0.8$ N.

Milli-equivalents of NaOH = $N \times V = 0.8 \times 20 = 16$

Milli-equivalents of $H_2SO_4$ = $0.8 \times 25 = 20$

The acid is in excess: excess meq of acid = $20 - 16 = 4$

Total volume = $20 + 25 = 45$ mL

$$N_{\text{resultant}} = \frac{\text{excess meq}}{\text{total volume}} = \frac{4}{45} = 0.0889 \text{ N}$$

The resulting solution is acidic with normality ≈ 0.089 N (about N/11).

(a) 1

The thiosulphate ion is $S_2O_3^{2-}$ (valency 2). In $M_2S_2O_3$, two M atoms balance a charge of −2, so each M is $M^+$. Its valency is 1 (as in $Na_2S_2O_3$).

(b) 0.5M

$KCl$ does not react with $NaNO_3$. It only doubles the volume. Mixing equal volumes halves the concentration of $NO_3^-$: $M = \frac{1 \times V}{2V} = 0.5$ M.

a) i) Primary standard solution: a solution of exactly known concentration made by dissolving a known mass of a primary standard substance (very pure, stable, non-hygroscopic, not changed by air, of fixed composition and high equivalent weight) in water and making the volume exactly up to the mark. Example: oxalic acid ($H_2C_2O_4\cdot2H_2O$) or anhydrous $Na_2CO_3$ solution.

ii) Indicator: no external indicator is needed. $KMnO_4$ itself acts as the indicator (self-indicator). Its purple colour disappears while oxalic acid remains. The first permanent light pink colour from one extra drop of $KMnO_4$ marks the end point.

iii) Why warmed with dil. $H_2SO_4$:

• Dilute $H_2SO_4$ provides the acidic medium so that $MnO_4^-$ is reduced to colourless $Mn^{2+}$ and brown $MnO_2$ does not form.
• The reaction is very slow at room temperature, so it is warmed to about 60–70 °C to make it fast enough. ($Mn^{2+}$ formed later catalyses the reaction.)

$$2KMnO_4 + 3H_2SO_4 + 5H_2C_2O_4 \rightarrow K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O$$

iv) Application in quality control: titration is used to find the strength or purity of products, for example the acetic acid content of vinegar, the acidity of milk, the amount of alkali in antacid tablets, or the percentage purity of raw chemicals and medicines before they are released.

b) Molarity and molality of the dibasic acid

Given: mass of acid = 17.7 g per litre of solution, molar mass = 118 g/mol, density = 1.0077 g/mL (the "gm/litre" in the question must mean g/mL, because 1 L of solution weighs about 1 kg).

i) Molarity

$$M = \frac{\text{mass}/\text{molar mass}}{\text{volume in L}} = \frac{17.7/118}{1} = 0.15 \text{ M}$$

ii) Molality

Mass of 1 L (1000 mL) of solution = $1000 \times 1.0077 = 1007.7$ g

Mass of solvent (water) = $1007.7 - 17.7 = 990$ g = 0.990 kg

$$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.15}{0.990} = 0.1515 \text{ mol/kg}$$

Molarity = 0.15 M; molality ≈ 0.152 m.

(d) 4N

For $Na_2CO_3$ the n-factor (total charge on the cation) is 2, so $N = M \times n = 2 \times 2 = 4$ N.

(b) 0.2 M

For $Na_2CO_3$, n-factor = 2, so $M = \dfrac{N}{n} = \dfrac{0.4}{2} = 0.2$ M.

a) Normality equation

Deduction of the normality equation $N_1V_1 = N_2V_2$

Suppose $V_1$ mL of an acid of normality $N_1$ is exactly neutralised by $V_2$ mL of a base of normality $N_2$.

• Normality = number of gram equivalents of solute dissolved per litre of solution.
• 1000 mL of a 1 N solution contains 1 g equivalent, so

$$\text{g-equivalents of acid in } V_1 \text{ mL} = \frac{N_1V_1}{1000}$$ $$\text{g-equivalents of base in } V_2 \text{ mL} = \frac{N_2V_2}{1000}$$

• By the law of equivalence, substances react in the ratio of their equivalents, so at the end point

$$\frac{N_1V_1}{1000} = \frac{N_2V_2}{1000}$$ $$\therefore\; N_1V_1 = N_2V_2$$

This is the normality equation (acid side = base side, or oxidant side = reductant side).

b) Reasons

i) Methyl orange is used because $Na_2CO_3$ (a weak base, salt of the weak acid $H_2CO_3$) is titrated with the strong acid $H_2SO_4$:

$$Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2$$

At the equivalence point the solution holds $CO_2$/$H_2CO_3$ and is slightly acidic (pH about 4). Methyl orange changes colour in the pH range 3.1–4.4 (yellow in alkali, red in acid), so it shows the true end point: yellow turns orange-red. Phenolphthalein is not suitable: it changes colour at pH 8–10, which is reached when only half the carbonate has been converted to $NaHCO_3$, so it would show the end point too early.

ii) Oxalic acid is warmed (to about 60–70 °C) with dilute $H_2SO_4$ before titration with $KMnO_4$ because:

• Dilute $H_2SO_4$ gives the acidic medium needed for $MnO_4^-$ to be reduced fully to colourless $Mn^{2+}$ (5 electrons gained, equivalent weight of $KMnO_4$ = M/5 = 31.6). Without enough acid, brown $MnO_2$ forms and the end point is wrong. $H_2SO_4$ is used (not HCl, which $KMnO_4$ oxidises to $Cl_2$; not $HNO_3$, which is itself an oxidising agent).
• Warming speeds up the reaction, which is very slow at room temperature at the start. (Once some $Mn^{2+}$ forms it catalyses the reaction: autocatalysis.)

$$2KMnO_4 + 3H_2SO_4 + 5H_2C_2O_4 \rightarrow K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O$$

c) Value of x in $Na_2CO_3\cdot xH_2O$

Given: mass of sample = 0.70 g in 100 mL; 20 mL of this solution needs 19.8 mL of N/10 HCl.

Normality of the carbonate solution ($N_1V_1 = N_2V_2$):

$$N_1 \times 20 = 0.1 \times 19.8 \quad\Rightarrow\quad N_1 = 0.099 \text{ N}$$

g-equivalents in 100 mL:

$$\frac{0.099 \times 100}{1000} = 0.0099 \text{ g eq}$$

Equivalent weight of the hydrated salt:

$$E = \frac{\text{mass}}{\text{g eq}} = \frac{0.70}{0.0099} = 70.7$$

Molar mass = $E \times 2 = 141.4$ (n-factor of $Na_2CO_3$ = 2)

Molar mass of $Na_2CO_3\cdot xH_2O = 106 + 18x$:

$$106 + 18x = 141.4 \quad\Rightarrow\quad x = \frac{35.4}{18} = 1.97 \approx 2$$

x = 2 (the formula is $Na_2CO_3\cdot 2H_2O$).

(d) $\mathrm{AgNO_3}$

Chloride is found by precipitation (argentometric) titration. Standard $AgNO_3$ precipitates $Cl^-$ as white $AgCl$ ($Ag^+ + Cl^- \rightarrow AgCl\downarrow$), with $K_2CrO_4$ as indicator (Mohr's method). The end point is a red-brown $Ag_2CrO_4$ colour.

Deduction of the normality equation $N_1V_1 = N_2V_2$

Suppose $V_1$ mL of an acid of normality $N_1$ is exactly neutralised by $V_2$ mL of a base of normality $N_2$.

• Normality = number of gram equivalents of solute dissolved per litre of solution.
• 1000 mL of a 1 N solution contains 1 g equivalent, so

$$\text{g-equivalents of acid in } V_1 \text{ mL} = \frac{N_1V_1}{1000}$$ $$\text{g-equivalents of base in } V_2 \text{ mL} = \frac{N_2V_2}{1000}$$

• By the law of equivalence, substances react in the ratio of their equivalents, so at the end point

$$\frac{N_1V_1}{1000} = \frac{N_2V_2}{1000}$$ $$\therefore\; N_1V_1 = N_2V_2$$

This is the normality equation (acid side = base side, or oxidant side = reductant side).

(a) Name of the titration: redox (oxidation–reduction) titration, specifically permanganometric titration. $KMnO_4$ is the oxidising agent and oxalic acid the reducing agent.

(b) Equivalent weight of $KMnO_4$ (in acidic medium):

$$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$

Mn changes from +7 to +2 (5 electrons gained). Molar mass of $KMnO_4$ = 39 + 55 + 64 = 158.

$$E = \frac{158}{5} = 31.6$$

(c) Normality of $KMnO_4$

Concordant reading = 11.0 mL.

Oxalic acid: $N_1 = 0.1$ N (decinormal), $V_1 = 10$ mL. $KMnO_4$: $V_2 = 11.0$ mL, $N_2 = ?$

$$N_1V_1 = N_2V_2 \;\Rightarrow\; 0.1 \times 10 = N_2 \times 11.0$$ $$N_2 = \frac{1}{11} = 0.0909 \text{ N}$$

Normality of $KMnO_4$ ≈ 0.091 N.

(d) Why dil. $H_2SO_4$ is added: it gives the acidic medium in which $MnO_4^-$ is reduced completely to colourless $Mn^{2+}$ (5 e⁻ change). Without enough acid, brown $MnO_2$ precipitates (only 3 e⁻ change), the colour change is masked and the result is wrong. HCl is not used because $KMnO_4$ oxidises it to $Cl_2$, and $HNO_3$ is not used because it is itself an oxidising agent. (The mixture is also warmed to 60–70 °C to speed up the slow reaction.)

(e) Titrant and titrand:

• Titrant (in the burette): $KMnO_4$ solution.
• Titrand / analyte (in the conical flask): standard oxalic acid solution acidified with dil. $H_2SO_4$.

(b) 0.1 N

0.5 M $H_2SO_4$ = $0.5 \times 2 = 1$ N. On dilution, $N_1V_1 = N_2V_2$: $1 \times 1 = N_2 \times 10$, so $N_2 = 0.1$ N.

(a) Name of the titration: acid–base (neutralisation) titration. Here the strength of an acid ($H_2SO_4$) is found with the help of a standard alkali ($Na_2CO_3$), so it is alkalimetry.

$$Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2$$

(b) Indicator and reason: Methyl orange is used because $Na_2CO_3$ (a weak base, salt of the weak acid $H_2CO_3$) is titrated with the strong acid $H_2SO_4$:

$$Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2$$

At the equivalence point the solution holds $CO_2$/$H_2CO_3$ and is slightly acidic (pH about 4). Methyl orange changes colour in the pH range 3.1–4.4 (yellow in alkali, red in acid), so it shows the true end point: yellow turns orange-red. Phenolphthalein is not suitable: it changes colour at pH 8–10, which is reached when only half the carbonate has been converted to $NaHCO_3$, so it would show the end point too early.

(c) Normality of $H_2SO_4$

Concordant volume of $H_2SO_4$ = 20.0 mL.

$Na_2CO_3$: $N_1 = 0.1$ N, $V_1 = 25$ mL. $H_2SO_4$: $V_2 = 20.0$ mL, $N_2 = ?$

$$N_1V_1 = N_2V_2 \;\Rightarrow\; 0.1 \times 25 = N_2 \times 20$$ $$N_2 = \frac{2.5}{20} = 0.125 \text{ N}$$

Normality of $H_2SO_4$ = 0.125 N.

(d) 2 g

g-equivalents of HCl = $\dfrac{2 \times 20}{1000} = 0.04$. Equivalent weight of $CaCO_3$ = 100/2 = 50, so mass = $0.04 \times 50 = 2$ g.

This is a back titration. The metal reacts with part of the acid, and the leftover acid is found with NaOH.

$$M + H_2SO_4 \rightarrow MSO_4 + H_2$$

Given:

• Mass of divalent metal = 4.5 g
• $H_2SO_4$: 100 mL of 2 M, normality factor f = 1.01
• Unreacted acid needed 30 mL of 1 N NaOH

Step 1: Actual normality of the acid

2 M $H_2SO_4$ = $2 \times 2 = 4$ N (approx.)

$$N_{\text{actual}} = f \times N_{\text{approx}} = 1.01 \times 4 = 4.04 \text{ N}$$

Step 2: Total acid taken

$$\text{meq of } H_2SO_4 = 4.04 \times 100 = 404 \text{ meq}$$

Step 3: Unreacted acid

$$\text{meq of NaOH} = 1 \times 30 = 30 \text{ meq} = \text{meq of unreacted acid}$$

Step 4: Acid that reacted with the metal

$$404 - 30 = 374 \text{ meq} = 0.374 \text{ g eq}$$

By the law of equivalence, g-eq of metal = 0.374.

Step 5: Equivalent weight and atomic mass

$$E = \frac{\text{mass}}{\text{g eq}} = \frac{4.5}{0.374} = 12.03$$ $$\text{Atomic mass} = E \times \text{valency} = 12.03 \times 2 \approx 24.06$$

Atomic mass of the metal ≈ 24 amu (the metal is magnesium).

Normality (N): the number of gram equivalents of solute dissolved per litre of solution.

$$N = \frac{\text{g-equivalents of solute}}{\text{volume of solution in L}} = \frac{w}{E \times V}$$

Molarity (M): the number of moles of solute dissolved per litre of solution.

$$M = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{w}{M_w \times V}$$

(where $w$ = mass of solute in g, $E$ = equivalent weight, $M_w$ = molar mass, $V$ = volume in L)

Relation:

Equivalent weight is related to molar mass by

$$E = \frac{M_w}{n}$$

where $n$ is the n-factor (basicity of an acid, acidity of a base, total positive valency of a salt, or number of electrons gained/lost by a redox reagent).

Substituting in the expression for normality:

$$N = \frac{w}{\frac{M_w}{n} \times V} = n \times \frac{w}{M_w \times V}$$ $$\therefore\; N = n \times M$$

Normality = n-factor × molarity. For example, 1 M $H_2SO_4$ (n = 2) = 2 N; 1 M HCl (n = 1) = 1 N; 0.1 M $KMnO_4$ in acid medium (n = 5) = 0.5 N.

(a) 0.94 N

5% (w/v) means 5 g in 100 mL = 50 g/L. Equivalent weight of $Na_2CO_3$ = 106/2 = 53, so $N = \dfrac{50}{53} = 0.943$ N ≈ 0.94 N.

a) Primary vs secondary standard substances

b) i) Normality of oxalic acid

Molar mass of $C_2H_2O_4\cdot2H_2O$ = 126, basicity = 2, so equivalent weight = 63.

$$N = \frac{w}{E \times V(\text{L})} = \frac{3.15}{63 \times 0.5} = 0.1 \text{ N}$$

Normality of oxalic acid = 0.1 N (decinormal).

ii) Mass of $KMnO_4$ per litre

The whole 500 mL of acid was oxidised by 400 mL of $KMnO_4$ (the permanent pink colour is the end point).

$$N_1V_1 = N_2V_2 \;\Rightarrow\; 0.1 \times 500 = N_2 \times 400$$ $$N_2 = 0.125 \text{ N}$$

Strength (g/L) = normality × equivalent weight:

$$S = 0.125 \times 31.5 = 3.94 \text{ g/L}$$

About 3.94 g of $KMnO_4$ is present in 1 litre of the solution.

(a) methyl orange

$H_2SO_4$ is a strong acid and $Na_2CO_3$ is a weak base, so the equivalence point lies on the acidic side (pH about 4, because of the $H_2CO_3$ formed). Methyl orange (pH range 3.1–4.4) changes colour sharply there and reflects the full reaction $Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2$. Phenolphthalein (pH 8.3–10) would mark only the half-way stage ($CO_3^{2-} \rightarrow HCO_3^-$), and litmus gives no sharp end point.

(b) Phenolphthalein

Ethanoic acid is a weak acid and NaOH is a strong base. At the equivalence point the solution contains sodium ethanoate, which hydrolyses to give a basic solution (pH about 8.5–9). Phenolphthalein changes colour in the range pH 8.3–10, which covers this point. Methyl orange and methyl red change in the acidic range and would show the end point too early.

a) Law of chemical equivalence [1]

Substances react with each other in the ratio of their equivalent weights; equal numbers of gram-equivalents of the reactants react completely, and also give equal gram-equivalents of products. In titration, at the end point the gram-equivalents of the acid equal those of the base.

b) Normality equation [2]

Gram-equivalents of a solute in $V$ mL of solution of normality $N$:

$$\text{g eq} = \frac{N \times V}{1000}$$

By the law of equivalence, at the end point:

$$\frac{N_1V_1}{1000} = \frac{N_2V_2}{1000} \Rightarrow N_1V_1 = N_2V_2$$

c) Normality of the mixture [2]

Milli-equivalents of each solution (these are all acids, so they simply add):

• HCl: $10 \times \dfrac{1}{2} = 5$
• $HNO_3$: $30 \times \dfrac{1}{10} = 3$
• $H_2SO_4$: $60 \times \dfrac{1}{5} = 12$

Total = $5 + 3 + 12 = 20$ meq. Total volume = $10 + 30 + 60 = 100$ cc.

$$N_{mix} = \frac{20}{100} = 0.2\ N$$

The normality of the mixture is N/5 (0.2 N).

(a) Sodium carbonate

A primary standard must be very pure, stable, non-hygroscopic and of definite composition. Anhydrous $Na_2CO_3$ meets these needs. $KMnO_4$, $FeSO_4$ (oxidised by air) and NaOH (absorbs moisture and $CO_2$) are secondary standards.

(a) The titration is a redox titration (permanganometry). A redox titration is one in which the substance in the burette oxidises or reduces the substance in the flask, so the end point is found from the transfer of electrons between oxidising and reducing agents. Here $KMnO_4$ (oxidising agent) oxidises oxalic acid (reducing agent) in acidic medium.

(b) $KMnO_4$ is not a primary standard because it is not available in pure form (it always contains $MnO_2$ impurity), and it is slowly decomposed by light, heat and traces of organic matter, so its composition and strength change on standing.

(c) In acidic medium: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$, so n-factor = 5.

$$\text{Equivalent mass} = \frac{158}{5} = 31.6$$

(d) No external indicator is used. $KMnO_4$ acts as a self-indicator: its purple colour changes to almost colourless $Mn^{2+}$ as it is added. At the end point, one extra drop gives a permanent light pink colour, which shows the reaction is complete.

a)

i) Equivalent weight: the mass of a substance that combines with or displaces 1 part by mass of hydrogen (or 8 parts of oxygen, or 35.5 parts of chlorine); equivalently, molar mass divided by the n-factor (valency or electrons exchanged).

ii) Standard solution: a solution whose concentration is accurately known. It is made by dissolving an accurately weighed primary standard in a known volume of solvent, or by standardising against such a solution.

iii) Indicator: a substance that shows the end point of a titration by a sharp, visible change (usually colour) at or near the equivalence point, e.g. methyl orange, phenolphthalein.

iv) Molality: the number of moles of solute dissolved in 1 kg of solvent (unit mol kg$^{-1}$).

b) Let $x$ L of $N/2$ HCl be mixed with $(2 - x)$ L of $N/10$ HCl to give 2 L of $N/5$ HCl. Equivalents are conserved:

$$0.5\,x + 0.1\,(2 - x) = 0.2 \times 2$$ $$0.5x + 0.2 - 0.1x = 0.4$$ $$0.4x = 0.2 \Rightarrow x = 0.5\ \text{L}$$

So the volume of $N/2$ HCl = 0.5 L (500 mL) and of $N/10$ HCl = $2 - 0.5 =$ 1.5 L (1500 mL).

Check: $0.5 \times 0.5 + 1.5 \times 0.1 = 0.25 + 0.15 = 0.40$ eq $= 0.2 \times 2$. Correct.

(c) M/2

$H_3PO_3$ is dibasic: only two of its H atoms are attached to O (P–OH) and can be replaced. In this reaction 2 $H^+$ are replaced to form $Na_2HPO_3$, so the n-factor = 2 and

$$E = \frac{M}{2}$$

A) End point and primary standard

• Detection of completion: a few drops of methyl orange are added to the $Na_2CO_3$ solution in the conical flask. It is yellow in the alkaline solution. When the reaction is complete, one extra drop of acid turns it orange-red (pink). This is the end point. Methyl orange is chosen because the equivalence point of this weak base–strong acid titration is acidic (pH about 4).

$$Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2$$

• Yes, anhydrous $Na_2CO_3$ is a primary standard. It can be obtained in a very pure state, it is stable (does not decompose, is not oxidised by air, does not absorb $CO_2$), it has a fixed composition and a fairly high equivalent weight (53), and it dissolves readily in water. So a solution of exactly known strength can be made by direct weighing.

B) Extra $H_2SO_4$ required

Given: mass of $Na_2CO_3$ = 2.16 g, equivalent weight of $Na_2CO_3$ = 106/2 = 53; volume of $H_2SO_4$ = 400 mL, normality = 0.1 N; equivalent weight of $H_2SO_4$ = 98/2 = 49.

g-equivalents of $Na_2CO_3$:

$$\frac{2.16}{53} = 0.04075 \text{ g eq}$$

g-equivalents of $H_2SO_4$ present:

$$\frac{N \times V}{1000} = \frac{0.1 \times 400}{1000} = 0.0400 \text{ g eq}$$

$Na_2CO_3$ is in excess, so the resultant solution is alkaline. Excess $Na_2CO_3$:

$$0.04075 - 0.0400 = 0.00075 \text{ g eq}$$

This needs an equal number of equivalents of $H_2SO_4$:

$$\text{Mass of } H_2SO_4 = 0.00075 \times 49 \approx 0.037 \text{ g}$$

About 0.037 g (37 mg) more $H_2SO_4$ is required to neutralise the solution completely.

(a) 49

Cr changes from +6 in $K_2Cr_2O_7$ to +3 in $Cr_2(SO_4)_3$: a gain of 3 e⁻ per Cr, so 6 e⁻ per $K_2Cr_2O_7$. Molar mass = 2(39) + 2(52) + 7(16) = 294, so $E = 294/6 = 49$.

a) Meaning of 2 M HCl: 2 moles of HCl (2 × 36.5 = 73 g) are dissolved in enough water to make 1 litre of solution.

b) Molarity vs molality

c) Significance of normality factor (f):

$$f = \frac{\text{actual mass taken}}{\text{theoretical mass required}} = \frac{\text{actual normality}}{\text{desired normality}}$$

• It tells how close the prepared solution is to the required strength (f = 1 means exact; f > 1 more concentrated; f < 1 more dilute).
• It lets us use a solution made with a slightly different mass without remaking it, since actual normality = f × intended normality.
• It is used to calculate how much water to add (if f > 1) or solute to add (if f < 1) to bring the solution to exact strength.

d) Molarity of liquor ammonia

Given: 25% (w/w) $NH_3$, specific gravity (density) = 0.91 g/mL, molar mass of $NH_3$ = 17 g/mol.

Mass of 1 L (1000 mL) of solution = $1000 \times 0.91 = 910$ g

Mass of $NH_3$ in it = $\dfrac{25}{100} \times 910 = 227.5$ g

$$\text{Molarity} = \frac{227.5/17}{1 \text{ L}} = 13.38 \text{ M}$$

Molarity of liquor ammonia ≈ 13.4 M.

a) Normality of the mixture

Total milliequivalents of acid (meq = $N \times V$ in mL):

$$\text{HCl: } 0.5 \times 100 = 50 \qquad HNO_3: 0.1 \times 20 = 2 \qquad H_2SO_4: 0.2 \times 50 = 10$$

Total meq = $50 + 2 + 10 = 62$; total volume = $100 + 20 + 50 = 170$ mL.

$$N_{mix} = \frac{62}{170} = 0.365\ \text{N}$$

b) pH of the mixture

All three are strong acids, so each equivalent supplies one mole of $H^+$ and $[H^+] = 0.365$ M.

$$pH = -\log(0.365) = 0.44$$

c) Decinormal solution

A decinormal solution is one that contains $\dfrac{1}{10}$ gram equivalent (0.1 g-eq) of solute per litre of solution, i.e. its normality is $N/10 = 0.1$ N.

(a) A primary standard substance is a pure, stable compound that can be weighed accurately and used to prepare a solution of exactly known concentration directly. It is non-hygroscopic and has a high molar mass. Examples are oxalic acid and anhydrous $Na_2CO_3$.

(b) Near the equivalence point the pH of a weak acid - weak base titration changes very gradually, with no sharp jump. No ordinary indicator changes colour sharply at the end point, so the end point cannot be detected accurately.

(c) Actual normality of the oxalic acid $= 0.1 \times 1.01 = 0.101$ N.

By the law of equivalence, $N_1V_1 = N_2V_2$:

$$0.101 \times 10 = N_{NaOH} \times 10 \Rightarrow N_{NaOH} = 0.101 \text{ N}$$

Strength of NaOH $= 0.101 \times 40 = 4.04$ g $L^{-1}$, which is $0.404$ g per 100 mL.

$$x = 0.404\ \% \text{ (w/v)}$$