A sample of sodium hydrogen carbonate solid weighing 0.1015 g requires 47.21 mL

A sample of sodium hydrogen carbonate solid weighing 0.1015 g requires 47.21 mL

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September 3, 2023
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A sample of sodium hydrogen carbonate solid weighing 0.1015 g requires 47.21 mL of a hydrochloric acid solution to react completely.

Calculate the molarity of the hydrochloric acid solution.

Answer and ExplanationSolution by a verified expert

Explanation
Number of moles of sodium hydrogen carbonate (NaHCO3 ) reacted = Amount of sample/ Molar Mass of NaHCO3

Molar Mass of NaHCO3=84 g

Number of moles of sodium hydrogen carbonate (NaHCO3 ) reacted=0.1015 g/84 g

=0.0012083 moles

Balanced Equation for the reaction is

NaHCO3 (aq) + HCl (aq) → NaCl (aq) + CO2 (g) + H2O (l)

Here 1 mole of NaHCO3 reacts with 1 mole of HCl

Similarly 0.0012083 moles of NaHCO3 reacts with 0.0012083 mole of HCl

Therefore number of moles of HCl reacted =0.0012083 moles

Molarity of HCl= Number of moles / Volume of HCl in L

=0.0012083/0.04721 L

=0.02559 M

If you have any doubts ask in the comment section. Have a Good Day

Answer

Molarity of the hydrochloric acid solution = 0.02559 M

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