What is Water of Crystallization?
Water of crystallization is an impurity in a compound. It does not take part in the chemical reaction. Hence, concentrations calculated from results during titration is that of the anhydrous substance. Water of crystallization and the percentage of water of crystallization can be calculated during titration.
The molecules of water of crystallization contained in a hydrated salt are given by the following relations.
= \( \frac{Mass\; concentration \;of \; water \; molecules}{Molar\; mass \;of \; water \; molecules}\)
\( \frac{Mass\; concentration \;of \; anyhdrous\;salt}{Mass\; concentration \;of \; hydrated \;salt} = \frac{Molar\; mass \;of \; anyhdrous\;salt}{Molar\; mass \;of \; hydrated\;salt} \)
Mass concentration of water molecules =
Mass concentration of hydrated salt – Mass concentration of anhydrous salt.
Percentage (%) of water of crystallization =
\( \frac{Mass\; concentration \;of \; water \; \; of \; crystallization}{Mass\; concentration \;of \; hydrated \; salt} \; \times \; \frac{100}{1} \)Example 1
A is a solution containing 0.1 moldm-3 HNO3.
B is a solution of anhydrous Na2CO3 salt.
The equation for the reaction is thus:
2HNO3(aq) + Na2CO3(aq) → 2NaNO3(aq) + H2O(aq) + CO2(g)
Put A into the burette and titrate it against 20.0 cm3 or 25.0 cm3 portion of B using methyl orange indicator. Note the first burette reading. This is your rough titre. Repeat the titration three more times and record your results. Calculate the average volume of A used.
(b) From the result and information provided, calculate the:
i. Concentration of B in moldm-3;
ii. Concentration of B in gdm-3
(c) Given that B contains 3.60 g of the hydrated Na2CO3 salt in 500 cm3 of solution, calculate the percentage of water of crystallization of the hydrated salt.
(WAEC)
Results and Calculations
Indicator used: Methyl orange
The volume of pipette used: 25.00 cm3
The titre values obtained on performing the experiment is presented below.
| Titration | Rough | 1st | 2nd |
| Final burette reading (cm3) | 13.10 | 26.00 | 12.70 |
| Initial burette reading (cm3) | 0.00 | 13.10 | 0.00 |
| Volume of Acid used (cm3) | 13.10 | 12.90 | 12.70 |
Average titre value = \( \frac{12.90 \; + \; 12.70}{2} \\ = \scriptsize 12.80cm^3 \)
CA = 0.1 moldm-3
VA = 12.80 cm3
nA = 2
CB =?
VB = 25.00 cm3
nB = 1
b(i) = \( \frac {C_A V_A}{C_B V_B} = \frac{n_A}{n_B} \\ \scriptsize C_B = \normalsize \frac{C_A V_A n_B}{V_B n_A} \\ \scriptsize C_B = \normalsize \frac{0.1 \; \times \; 12.80 \; \times \; 1 }{25 \; \times \; 2} \\ \scriptsize = 0.0256 moldm^{-3} \)
(ii) The molar mass of Na2CO3 = (23 x 2) + (12 x 1) + (16 x 3) = 106 gmol-1.
Concentration of B in g/dm3 = concentration of B in moldm-3 x molar mass
= 0.0256 moldm-3 x 106 gdm-3
= 2.71 gdm-3
(c). 500 cm3 of the hydrated B contains 3.60 g
1000 cm3 will contain \( \frac {3.60}{500} \scriptsize \; \times \; 1000 cm^3 \\ \scriptsize = 7.20 gdm^{-3} \)
To find the number of water molecules in hydrated salt B,
\( \frac{Mass\; concentration \;of \; anyhdrous\;salt}{Mass\; concentration \;of \; hydrated \;salt} = \frac{Molar\; mass \;of \; anyhdrous\;salt}{Molar\; mass \;of \; hydrated\;salt} \)
\( \frac{106}{106 \; + \; 18.6} = \frac{2.74}{7.20} \\ \scriptsize 2.71(106 \; + \; 18x) = 106 \; \times \; 7.20 \\ \scriptsize 287.26 \; + \; 48.78x = 763.2 \\ \scriptsize 49.14x = 763.2 \; – \; 287.26 \\ \scriptsize 49.14x = 475.94 \\ \scriptsize x = \normalsize \frac{475.94}{49.14} \\ \scriptsize x = 9.76 \\ \scriptsize 9.76 \approx 10 \)Therefore, the number of water molecules in the salt is 10.
Evaluation Questions
1. The following table shows burette readings obtained when 25.0 cm3 of a solution of sodium trioxocarbonate (IV) containing 1.30 g in 250 cm3 of solution was titrated against a solution of hydrochloric acid of an unknown concentration:
| Burette reading | Rough | I | II |
| Final reading(cm3) | 23.50 | 44.20 | 25.20 |
| Initial reading(cm3) | 2.00 | 23.50 | 4.30 |
| Volume of acid used (cm3) |
a. Copy and complete the table.
b. Calculate the average volume of acid used.
c. Write a balanced equation for the reaction in the titration
i. From the information obtained in 1(a) above, calculate the:
ii. concentration of Na2CO3 in moldm-3;
iii. concentration of HCl in
iv. moldm-3 ;
v. gdm-3.
[H = 1.00; O = 16.0; Na = 23.0; Cl = 35.5] (WAEC, 2012)
View Solution2. In a titration experiment, 22.50 cm3 of an acid solution A containing 10.6 g of NaHSO4 per dm3 reacted with 25.0 cm3 of solution B containing X g of NaOH per dm3. The equation for the reaction is:
NaHSO4(aq) + NaOH(aq) → Na2SO4(aq) + H2O(l)
(a) From the information given above, calculate the:
(i) concentrate of A in moldm– 3;
(ii) concentrate of B in moldm– 3;
(iii) value of X;
(iv) mass of Na2SO4 formed during the reaction.
[H = 1.00; 0 = 16.0; Na = 23.0; S = 32.0] (WAEC, 2014)
(b)(i) Name a suitable indicator for the titration experiment.
(ii) State the apparatus used to measure the volume of solution: I. A; II B
(3). Briefly explain how titration is carried out.
View Solution
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