Answer:
Assuming 20.0 mL of 0.2 M NaOH is used to neutralize 30 mL of the HCl of unknown concentration, the concentration of the hydrochloric acid is 0.133 M.
Note: The volume of NaOH used isnot given and issp to be determined by titration.
Step-by-step explanation:
The question is a lab activity. Therefore, the volume of 0.2 M NaOH required for the neutralization of 30.0 mL of hydrochloric acid of unknown concentration is to be determined from titration. However, assuming a certain volume of the base is used for the complete neutralization of the acid, the concentration of the acid can be calculated.
Assuming 20.0 mL of 0.2 M NaOH is used to neutralize 30 mL of the HCl of unknown concentration, the concentration of the hydrochloric acid can be calculated thus:
Step 1: Balanced chemical equation of reaction:
HCl + NaOH ---> NaCl + H₂O
From the equation of the reaction, 1 mole of NaOH reacts with 1 mole of HCl to produce 1 mole of NaCl and water.
Step 2: Determine the number of moles of NaOH in 20.0 mL of 0.200 M solution
Molarity = number of moles/volume in Litres
Number of moles = molarity × volume in litres
Volume of NaOH in litres = 20 ml × 1 litre/1000ml = 0.02 litres
Number of moles NaOH = 0.02 L × 0.2 M = 0.004 moles
Therefore, 0.00 M of NaOH reacted with 0.004 moles of HCl
Step 3: Determine concentration of HCl
Molarity or concentration = number of moles / volume in litres
Number of moles of HCl = 0.004 moles
Volume of HCl in litres = 30.0 mL × 1 L/1000 mL = 0.03 L
Concentration of HCl = 0.004 moles / 0.03 L = 0.133 M
Therefore, assuming 20.0 mL of 0.2 M NaOH is used to neutralize 30 mL of the HCl of unknown concentration, the concentration of the hydrochloric acid is 0.133 M.