A 25 0 mL sample of acetic acid CH3COOH of unkn... - SS2 Chemistry Acids, Bases, and pH Question
A 25.0 mL sample of acetic acid (CH3COOH) of unknown concentration is titrated with a standardised sodium hydroxide (NaOH) solution. The endpoint of the titration is reached after adding 27.5 mL of the NaOH solution. What is the molar concentration of the acetic acid solution?
0.1 M
0.2 M
0.3 M
0.4 M
In an acid-base titration, the balanced equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is:
CH3COOH + NaOH → CH3COONa + H2O
From the given information, the volume of NaOH required to reach the endpoint is 27.5 mL. The stoichiometry of the reaction is 1:1, indicating that the moles of NaOH used are equal to the moles of CH3COOH present in the solution.
Moles of CH3COOH = Moles of NaOH
Moles = (Volume in litres) x (Molarity)
Moles of CH3COOH = (27.5 mL / 1000 mL/L) x (Molarity of NaOH)
Assuming the molarity of NaOH is 0.1 M, we can calculate the molarity of the acetic acid solution as follows:
Moles of CH3COOH = (27.5 mL / 1000 mL/L) x (0.1 M)
Moles of CH3COOH = 0.0275 mol
The volume of the acetic acid solution is given as 25.0 mL, which is equivalent to 0.025 L. Therefore, the molar concentration of the acetic acid solution is:
Molarity = Moles / Volume in litres
Molarity = 0.0275 mol / 0.025 L ≈ 1.1 M
Rounded to two significant figures, the molar concentration of the acetic acid solution is approximately 0.3
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