Question

determine strength and molarity of hcl solutions by titrating it against M/20 standard solution of Na2CO3​

Answer 1

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The molarity of hydrochloric acid is determined by titrating it against the standard solution of sodium carbonate using methyl orange as indicator. 2. Strength of the acid is determined by multiplying its molarity with its molecular mass which is 36.5.

Answer 2

Given: M/20 standard solution of $Na_{2} CO_{3}$ which means that the volume of $Na_{2} CO_{3}$ solution is 20 ml

Molarity of $Na_{2} CO_{3}$ = 0.05M

To Find: Strength of HCl solution

Molarity of HCl solution

Solution:

• Let the volume of the HCl solution be x ml
• Molarity of Na2co3 solution = 0.05M
• Volume of Na2co3 solution for neutralizing HCl solution = 20ml
• By using the formula mentioned below, the molarity of the HCl solution can be predicted,
• The product of molarity and volume of a compound is equal to the stoichiometric coefficient of the solution
• Let the molarity of $Na_{2} CO_{3}$ is $M_{2}$
• The volume of $Na_{2} CO_{3}$ is $V_{2}$

• The molarity of HCl is $M_{1}$
• The volume of HCl is $V_{1}$
• $M_{1} V_{1}$ = stoichiometric coefficient of HCl  ...(i)
• $M_{2} V_{2}$ = stoichiometric coefficient of $Na_{2} CO_{3}$       ...(ii)
• By dividing (i) with (ii) we get,
• = $\frac{M_{1} V_{1} }{M_{2} V_{2} } = \frac{stoichiometric coefficient of HCl}{stoichiometric coefficient of Na_{2} CO_{3} }$
• = $\frac{(M_{1}) (x)}{(0.05) (20)} = \frac{2}{1}$
• = $M_{1} = \frac{(0.05) (20) (2)}{x}$
• = $M_{1} = \frac{2}{x}$
• The strength of HCl can be calculated by the product of molar mass of HCl and the molarity of HCl.
• = Molar mass of HCl is 1 + 35.5 = 36.5 g/mol
• Strength = $\frac{2}{x}$ × 36.5 = $\frac{73}{x}$ g/L.

Hence, the strength of HCl is $\frac{73}{x} g/L$.