A flask contains 0.528 g of acid and a few drops of phenolphthalein indicator dissolved in water. The buret contains 0.200 M NaOH. 32.5 mL of the NaOH is titrated into the flask.
What is the molar mass of the acid (assuming it is diprotic and that the end point corresponds to the second equivalence point)?
Please explain.
To determine the molar mass of the acid, we need to use the concept of titration and the stoichiometry of the reaction between the acid and the sodium hydroxide (NaOH).
In this case, we are given the concentration of the NaOH solution (0.200 M) and the volume of NaOH solution required to reach the equivalence point in the titration (32.5 mL). The reaction between the acid and NaOH can be represented by the balanced chemical equation:
Acid (H2A) + 2NaOH -> Na2A + 2H2O
From the equation, we can see that the mole ratio between the acid and NaOH is 1:2. This means that 1 mole of acid reacts with 2 moles of NaOH.
First, let's calculate the number of moles of NaOH used in the titration:
moles of NaOH = volume of NaOH solution (L) × concentration of NaOH (mol/L)
Convert the volume of NaOH solution from mL to L:
volume of NaOH solution (L) = 32.5 mL ÷ 1000 mL/L = 0.0325 L
Now we can calculate the moles of NaOH:
moles of NaOH = 0.0325 L × 0.200 mol/L = 0.0065 mol
Since the mole ratio between the acid and NaOH is 1:2, the number of moles of the acid will be half of the moles of NaOH used in the reaction:
moles of acid = 0.0065 mol ÷ 2 = 0.00325 mol
Next, we need to find the mass of the acid:
mass of acid = moles of acid × molar mass of acid
Rearranging the formula, we can solve for the molar mass of the acid:
molar mass of acid = mass of acid ÷ moles of acid
The mass of the acid is given in the question as 0.528 g:
molar mass of acid = 0.528 g ÷ 0.00325 mol ≈ 162.46 g/mol
So, the molar mass of the acid, assuming it is diprotic and that the end point corresponds to the second equivalence point, is approximately 162.46 g/mol.