**Determine volumetrically, the value of ‘n’ i.e., the number of water molecules of crystallisation in washing soda \(Na_{2}CO_{3}.nH_{2}O\), 7.0 g of which have been dissolved per litre of the given solution. Provided \(\frac{M}{10} HCl\)**

**Chemical Equation:**

**Indicator.** Methyl orange.

**End Point.** Yellow to light pink (Acid in burette)

**Procedure**

- Rinse and fill the burette with the standard HCl solution.
- Rinse the pipette with the sodium carbonate solution and pipette out 20 ml of this solution in the titration flask.
- Add 2-3 drops of methyl orange indicator to the titration flask. The colour of the solution becomes yellow.
- Note the initial reading of the burette and run acid solution slowly in the titration flask till the solution becomes light pink.
- Note the final reading of the burette and find out the volume of acid used.
- Repeat the procedure 4-5 times to get a set of at least three concordant readings.

**Observations**

Molarity of given HCl solution = \(\frac{M}{10}\)

Volume of sodium carbonate solution taken for each titration = 20.0 ml

S.No. |
Initial reading of the burette |
Final reading of the burette |
Volume of the acid solution used |

1. | — | _ |
— ml |

2. | — | — | — ml |

3. | — | — | — ml |

4. | — | — | — ml |

Concordant volume = x ml (say).

Calculations

The molarity of standard HCl = 0.1 M

Since in the balanced equation two moles of HC1 react with one mole of Na_{2}C0_{3}.nH_{2}0.

\(\frac { 0.1\times x }{ { M }_{ { Na }_{ 2 }{ CO }_{ 3 } }\times 20.0 } =\frac { 2 }{ 1 } \)

\({ M }_{ { Na }_{ 2 }{ CO }_{ 3 } }=\frac { 1\times x\times 0.1 }{ 2\times 20.0 } =\frac { x }{ 400 } \)

\({ M }_{ { Na }_{ 2 }{ CO }_{ 3 } }=\frac { Streanth\quad per\quad litre }{ Molar\quad mass\quad of\quad { Na }_{ 2 }{ CO }_{ 3 }.n{ H }_{ 2 }O } \)

\(Molar\quad mass\quad of\quad { Na }_{ 2 }{ CO }_{ 3 }.n{ H }_{ 2 }O=\frac { Streanth\quad per\quad litre }{ { M }_{ { Na }_{ 2 }{ CO }_{ 3 } } } \)

\(=\frac { 7.0 }{ \frac { x }{ 400 } } \quad g{ mol }^{ -1 }\)

But molar mass of hydrated sodium carbonate = (106 + 18n) g mol^{-1}

therefore,

\(\frac { 7.0 }{ \frac { x }{ 400 } } =108+18n\)Knowing the titre value, x, the value of n can be calculated.

**Result**

The number of the molecules of water of crystallisation in washing soda is …..

Note. The result is expressed as nearest whole number. The value of n is 10 in this case.

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