Titrating Sodium Carbonate with Hydrochloric Acid – Concentration of Solutions (Senior 3 Chemistry) | Ganzaa

Concentration of Solutions

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Senior 3 · Chemistry

Concentration of Solutions

5 lessons

Lesson 5 of 5

Titrating Sodium Carbonate with Hydrochloric Acid

25 min

Crystals that lose their water

Washing soda, which you met in Unit 1 as a water softener, is sold as clear crystals of sodium carbonate-10-water, $\mathrm{Na_2CO_3{\cdot}10H_2O}$ . Left in an open jar, the crystals slowly lose water to the air and crumble to a white powder, so a gram of old washing soda contains more sodium carbonate than a gram of fresh crystals. A titration with hydrochloric acid measures exactly how much sodium carbonate a sample really holds, and from that, how much water of crystallisation it has left.

The reaction and its indicator

Sodium carbonate is a base that needs two moles of hydrochloric acid for each mole:

$\mathrm{Na_2CO_3(aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g)}$

The procedure is the one in Lesson 4: the carbonate solution is pipetted into the flask, the acid is in the burette, and the indicator is methyl orange. The solution starts yellow; near the end, bubbles of carbon dioxide appear where the acid lands; the end point is the first permanent orange (Figure 5.1).

Important

Why methyl orange, not phenolphthalein? The acid reacts with carbonate in two stages. First it turns carbonate into hydrogencarbonate, $\mathrm{CO_3^{2-} + H^+ \rightarrow HCO_3^-}$ ; then it turns hydrogencarbonate into carbon dioxide and water, $\mathrm{HCO_3^- + H^+ \rightarrow H_2O + CO_2}$ . Phenolphthalein loses its pink colour at the end of the first stage, when only half the acid has been added. Methyl orange changes at the end of the second stage, which matches the full 1 : 2 equation.

Flasks from carbonate titrations with two different indicators: phenolphthalein, pink (left), and methyl orange, orange (right). Photo: LLosss, CC BY 4.0, via Wikimedia Commons

Worked example

Problem (standardising an acid). $25.0\,\mathrm{cm^3}$ of the standard $0.100\,\mathrm{mol\,dm^{-3}}$ sodium carbonate solution from Lesson 3 needs a mean titre of $23.80\,\mathrm{cm^3}$ of hydrochloric acid. Find the concentration of the acid.

Solution. $n(\mathrm{Na_2CO_3}) = 0.100 \times 0.0250 = 0.00250\,\mathrm{mol}$

Ratio $\mathrm{Na_2CO_3 : HCl} = 1 : 2$ , so $n(\mathrm{HCl}) = 2 \times 0.00250 = 0.00500\,\mathrm{mol}$

$c(\mathrm{HCl}) = \frac{0.00500}{0.02380} = 0.210\,\mathrm{mol\,dm^{-3}}$

Finding the exact concentration of a solution in this way, against a standard, is called standardising it.

Finding the water of crystallisation

3.58 g of washing soda crystals, $\mathrm{Na_2CO_3{\cdot}xH_2O}$ , was dissolved and made up to $250\,\mathrm{cm^3}$ in a volumetric flask. Portions of $25.0\,\mathrm{cm^3}$ were titrated with $0.100\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid, using methyl orange.

Table 5.1: Titration results (illustrative)

Titration	rough	1	2	3
final reading (cm³)	25.40	25.25	25.45	25.10
initial reading (cm³)	0.00	0.20	0.50	0.10
titre (cm³)	25.40	25.05	24.95	25.00

Concordant titres: 25.05, 24.95 and 25.00 cm³; mean $= 25.00\,\mathrm{cm^3}$ . Figure 5.2 sets out the calculation.

Worked example

Problem. Use Table 5.1 to find $x$ in $\mathrm{Na_2CO_3{\cdot}xH_2O}$ . ( $\mathrm{Na_2CO_3} = 106$ , $\mathrm{H_2O} = 18$ )

Solution. $n(\mathrm{HCl}) = 0.100 \times 0.02500 = 0.00250\,\mathrm{mol}$ , so $n(\mathrm{Na_2CO_3}) = \frac{0.00250}{2} = 0.00125\,\mathrm{mol}$ in each 25.0 cm³ portion.

The flask held ten such portions: $0.00125 \times 10 = 0.0125\,\mathrm{mol}$ in the 3.58 g of crystals.

$M = \frac{3.58}{0.0125} = 286\,\mathrm{g\,mol^{-1}} \qquad x = \frac{286 - 106}{18} = 10$

The crystals are $\mathrm{Na_2CO_3{\cdot}10H_2O}$ : this sample had lost none of its water.

Worked example

Problem (percentage purity). 2.80 g of impure anhydrous sodium carbonate is made up to $250\,\mathrm{cm^3}$ . A $25.0\,\mathrm{cm^3}$ portion needs $22.60\,\mathrm{cm^3}$ of $0.200\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid. The impurities do not react with acid. Find the percentage purity.

Solution. $n(\mathrm{HCl}) = 0.200 \times 0.02260 = 0.00452\,\mathrm{mol}$ ; $n(\mathrm{Na_2CO_3}) = 0.00226\,\mathrm{mol}$ in 25.0 cm³, so $0.0226\,\mathrm{mol}$ in 250 cm³.

Mass of $\mathrm{Na_2CO_3} = 0.0226 \times 106 = 2.40\,\mathrm{g}$

$\text{percentage purity} = \frac{2.40}{2.80} \times 100 = 85.6\%$

Warning

Common mistakes and safety. (1) Using a 1 : 1 ratio: sodium carbonate needs two moles of HCl. (2) Forgetting to multiply by 10 when only a 25.0 cm³ portion of a 250 cm³ solution was titrated. (3) Stopping at the first flash of orange: swirl, and take the end point as the first permanent orange. (4) Wear eye protection; if acid or carbonate solution touches your skin or eyes, rinse with plenty of running water for at least 15 minutes.

Tip

Carbonate takes two. $\mathrm{Na_2CO_3 + 2HCl}$ : halve the moles of acid to get the moles of carbonate, or double the carbonate to get the acid. Methyl orange goes yellow, then orange at the end.

Free Chemistry Tutorials: Titration to Standardise a Hydrochloric Acid Solution. Watch a standard sodium carbonate solution being titrated with hydrochloric acid, and the colour of the indicator at the end point.

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Practice

Write the balanced equation, with state symbols, for the reaction of sodium carbonate solution with hydrochloric acid.
Why is methyl orange, and not phenolphthalein, used for this titration?
$25.0\,\mathrm{cm^3}$ of $0.050\,\mathrm{mol\,dm^{-3}}$ sodium carbonate needs $20.00\,\mathrm{cm^3}$ of hydrochloric acid. Find the concentration of the acid.
What volume of $0.100\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid reacts exactly with $25.0\,\mathrm{cm^3}$ of $0.080\,\mathrm{mol\,dm^{-3}}$ sodium carbonate?
$25.0\,\mathrm{cm^3}$ of a sodium carbonate solution needs $18.00\,\mathrm{cm^3}$ of $0.150\,\mathrm{mol\,dm^{-3}}$ HCl. Find its concentration in $\mathrm{mol\,dm^{-3}}$ and $\mathrm{g\,dm^{-3}}$ . ( $\mathrm{Na_2CO_3} = 106$ )
7.15 g of washing soda crystals, $\mathrm{Na_2CO_3{\cdot}xH_2O}$ , is made up to $250\,\mathrm{cm^3}$ . A $25.0\,\mathrm{cm^3}$ portion needs $25.00\,\mathrm{cm^3}$ of $0.200\,\mathrm{mol\,dm^{-3}}$ HCl. Find $x$ .
An old sample of washing soda has lost some water. Would the value of $x$ found by titration be higher or lower than 10? Explain.
A student titrates sodium carbonate with hydrochloric acid using phenolphthalein, and uses the 1 : 2 ratio in her calculation. Evaluate her answer for the concentration of the acid.

Show answersHide answers

$\mathrm{Na_2CO_3(aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g)}$
Phenolphthalein changes colour when only half the acid has been added (carbonate has become hydrogencarbonate). Methyl orange changes when all the carbonate has reacted, matching the 1 : 2 equation.
$n(\mathrm{Na_2CO_3}) = 0.050 \times 0.0250 = 0.00125\,\mathrm{mol}$ ; $n(\mathrm{HCl}) = 0.00250\,\mathrm{mol}$ ; $c = \frac{0.00250}{0.02000} = 0.125\,\mathrm{mol\,dm^{-3}}$ .
$n(\mathrm{Na_2CO_3}) = 0.080 \times 0.0250 = 0.0020\,\mathrm{mol}$ ; $n(\mathrm{HCl}) = 0.0040\,\mathrm{mol}$ ; $V = \frac{0.0040}{0.100} = 0.040\,\mathrm{dm^3} = 40.0\,\mathrm{cm^3}$ .
$n(\mathrm{HCl}) = 0.150 \times 0.01800 = 0.00270\,\mathrm{mol}$ ; $n(\mathrm{Na_2CO_3}) = 0.00135\,\mathrm{mol}$ ; $c = \frac{0.00135}{0.0250} = 0.0540\,\mathrm{mol\,dm^{-3}}$ ; $0.0540 \times 106 = 5.72\,\mathrm{g\,dm^{-3}}$ .
$n(\mathrm{HCl}) = 0.200 \times 0.02500 = 0.00500\,\mathrm{mol}$ ; $n(\mathrm{Na_2CO_3}) = 0.00250\,\mathrm{mol}$ per portion, $0.0250\,\mathrm{mol}$ in total; $M = \frac{7.15}{0.0250} = 286$ ; $x = \frac{286 - 106}{18} = 10$ .
Lower. With less water in each gram, the same mass contains more sodium carbonate, so the calculated molar mass is smaller and $x = \frac{M - 106}{18}$ comes out below 10.
Her answer is wrong, by a factor of two. With phenolphthalein the end point comes when only one mole of HCl has reacted per mole of carbonate, so the titre is about half the methyl-orange titre. Using 1 : 2 with that titre makes her acid concentration about twice the true value.

1Crystals that lose their water

2The reaction and its indicator

3Finding the water of crystallisation

Crystals that lose their water

The reaction and its indicator

Finding the water of crystallisation