Question

A sample of pure sodium carbonate 0.318 gram is dissolved in water and treated with HCl. A solution of volume of 60 ml is required to reach the methyl orange end point .Calculate the molarity of the acid

Answer 1

Solution

$$0.318g \quad 60ml$$ $$v = m/M = 0.318/106 = 0.003moles \quad x moles$$ $$Na_2CO_3 \quad + \quad 2HCl = 2NaCl + H_2O + CO_2$$ $$v = 1 mole \quad v = 2 moles$$ $$M = 2 \times 23 + 12 + 3 \times 16 = 106\,(g/mole) \quad M = 1 + 35,5 = 36,5\,(g/mole)$$

Proportionally

$$\frac{0.003moles}{1} = \frac{x}{2moles};$$ $$x = \frac{0.003 \times 2}{1} = 0,006\,moles;$$

Molarity of the acid:

$$\frac{0.006moles}{60ml} = \frac{x moles}{1000ml};$$ $$x = \frac{0.006 \times 1000}{60} = 0,1M.$$