If 0.314g sample of a solid monoprotic acid were dissolved in water and titrated to an end point that required 44.2 mL of 0.115M NaOH, what is the molecular mass of the unknown acid?
To determine the molecular mass of the unknown acid, we need to use the concept of stoichiometry and the titration equation.
The balanced chemical equation for the reaction between the acid and sodium hydroxide (NaOH) is:
acid + NaOH --> salt + water
From the equation, we can see that the acid reacts in a 1:1 ratio with NaOH. This means that one mole of acid reacts with one mole of NaOH.
Step 1: Calculate the number of moles of NaOH used in the titration.
The volume of NaOH used is 44.2 mL, which is equivalent to 0.0442 L. The concentration of NaOH solution is given as 0.115 M (moles per liter). Therefore, we can calculate the number of moles of NaOH using the formula:
moles of NaOH = concentration × volume
moles of NaOH = 0.115 M × 0.0442 L
Step 2: Calculate the number of moles of acid used in the titration.
Since the acid and NaOH react in a 1:1 ratio, the number of moles of acid used is the same as the number of moles of NaOH used.
moles of acid = moles of NaOH
Step 3: Calculate the molecular mass of the acid.
The molecular mass of a compound can be calculated using the formula:
molecular mass = mass / moles
In this case, the mass of the acid is given as 0.314 g.
molecular mass = 0.314 g / moles of acid
By substituting the value for the moles of acid from Step 2, we can calculate the molecular mass of the acid.
HA + NaOH ==> NaA + H2O
mols NaOH = M x L = ?
1 mol HA = 1 mol NaOH; therefore, mols HA = mols NaOH.
Then mol = g/molar mass. You know mpols and grams, solve for molar mass.