A sample of sodium hydrogen carbonate solid weighing 0.1015 g requires 47.21 mL
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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. |
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
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Answer
Molarity of the hydrochloric acid solution = 0.02559 M
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