f 10.0 mL of a solution of propanoic acid was titrated with 0.1063 mol/L sodium hydroxide, giving a titre of 18.2 mL, what is the concentration of the propanoic acid?Question 5Answer1.93e-4 mol/L0.193 mol/L0.232 mol/L1.93 mol/L0.0584 mol/L
Question
f 10.0 mL of a solution of propanoic acid was titrated with 0.1063 mol/L sodium hydroxide, giving a titre of 18.2 mL, what is the concentration of the propanoic acid?Question 5Answer1.93e-4 mol/L0.193 mol/L0.232 mol/L1.93 mol/L0.0584 mol/L
Solution
To solve this problem, we need to use the concept of stoichiometry in acid-base reactions. The reaction between propanoic acid and sodium hydroxide can be written as:
CH3CH2COOH + NaOH -> CH3CH2COONa + H2O
From this equation, we can see that one mole of propanoic acid reacts with one mole of sodium hydroxide.
First, we need to calculate the number of moles of sodium hydroxide used in the titration. This can be done using the formula:
moles = concentration * volume
where the concentration is 0.1063 mol/L and the volume is 18.2 mL (or 0.0182 L, because 1 mL = 0.001 L).
moles of NaOH = 0.1063 mol/L * 0.0182 L = 0.00193546 mol
Since the stoichiometry of the reaction is 1:1, the number of moles of propanoic acid is also 0.00193546 mol.
The concentration of propanoic acid can then be calculated using the formula:
concentration = moles / volume
where the volume is 10.0 mL (or 0.01 L).
concentration of propanoic acid = 0.00193546 mol / 0.01 L = 0.193546 mol/L
So, the concentration of the propanoic acid is approximately 0.193 mol/L. Therefore, the correct answer is 0.193 mol/L.
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