X is a solution containing 6.0g/dm³ of HCl. Y contains 12g/dm³ of Na2CO3.XH2O. A student treated 25cm³ of Y with x using methyl orange as an indicator, and the average titer value was 12.9cm³.
a) Write a balanced equation for the reaction.
b) Calculate:
i) the concentrated anhydrous Na2CO3 in solution Y
ii) the value of x
iii) the percentage of water of crystallization in solution y
c) What type of salt is NA2CO3.XH2O?
Answer (1)
Let's break down this problem step by step.
a) Balanced Equation
The reaction between hydrochloric acid (HCl) and sodium carbonate (Na₂CO₃ · xH₂O) can be written as:
2 HCl + Na 2 CO 3 → 2 NaCl + H 2 O + CO 2
b) Calculations
i) Concentration of Anhydrous Na₂CO₃ in Solution Y
Since Y contains 12 g/dm³ of Na₂CO₃ · xH₂O, we need to know the molar mass of Na₂CO₃ and Na₂CO₃ · xH₂O.
The molar mass of Na₂CO₃ is approximately 106 g/mol.
ii) Value of x
To find the value of x, we need to know that the number of moles of HCl equals the number of moles of Na₂CO₃:
Concentration of HCl: 6.0 g/dm³
Molar mass of HCl: approximately 36.5 g/mol
Number of moles of HCl in the volume of X used:
Moles of HCl = 36.5 g/mol 6.0 g/dm 3 × 12.9 × 1 0 − 3 dm 3
Number of moles of Na₂CO₃ in 25 cm³ (0.025 dm³):
Using the stoichiometry of the balanced equation, 2 moles of HCl react with 1 mole of Na₂CO₃:
Moles of Na 2 CO 3 = 2 Moles of HCl
iii) Percentage of Water of Crystallization in Solution Y
Now, to find the percentage of water in Na₂CO₃ · xH₂O, we can use the formula:
Percentage of water = ( Molar mass of Na 2 CO 3 ⋅ x H 2 O x × 18 g/mol ) × 100
Using these measurements, you'll solve for x and subsequently calculate the percentage.
c) Type of Salt
Na₂CO₃ · xH₂O is known as a hydrated salt. It's a type of inorganic salt that contains water molecules within its crystal structure.
By going through each step logically and completing the calculations based on given values, you'll arrive at the needed values for x and the concentrations.
