A 0.1873 g sample of a pure, solid acid, H2X was dissolved in water and titrated with 0.1052 M NaOH solution.
The balanced equation for the neutralization reaction occurring is
H2X(aq) + 2NaOH(aq) �¨ Na2X(aq) + 2H2O(l)
If the molar mass of H2X is 85.00 g/mol, calculate the volume of NaOH solution needed in the titration. (41.89 mL)
How do I solved for liters if nothing is given in the question?
To calculate the volume of NaOH solution needed in the titration, we need to use the concept of stoichiometry and the balanced equation of the reaction.
First, let's determine the number of moles of H2X in the 0.1873 g sample. We can use the molar mass of H2X (85.00 g/mol) to do this:
moles of H2X = mass / molar mass
moles of H2X = 0.1873 g / 85.00 g/mol
moles of H2X = 0.00220176 mol
From the balanced equation, we know that the stoichiometric ratio between H2X and NaOH is 1:2. This means that for every 1 mole of H2X, we need 2 moles of NaOH.
Therefore, the number of moles of NaOH needed to neutralize the H2X is:
moles of NaOH = 2 * moles of H2X
moles of NaOH = 2 * 0.00220176 mol
moles of NaOH = 0.00440352 mol
Now we can use the molarity of the NaOH solution to calculate the volume of NaOH solution needed. The relationship between moles, volume, and molarity is given by the equation:
moles = volume (in liters) * molarity
Rearranging the equation, we can solve for the volume:
volume (in liters) = moles / molarity
volume (in liters) = 0.00440352 mol / 0.1052 mol/L
volume (in liters) ≈ 0.04189 L
Finally, we convert the volume from liters to milliliters:
volume (in mL) = 0.04189 L * 1000 mL/L
volume (in mL) ≈ 41.89 mL
Therefore, the volume of NaOH solution needed in the titration is approximately 41.89 mL.
moles H2X present = 0.1873/85.00 = ??
moles NaOH = 2 x that number.
MNaOH = moles NaOH/L NaOH
Solve for L NaOH and convert to mL. 41.89 mL is the correct answer.