Answer:
The molar mass of the unknown acid is 89 g/mol
Explanation:
<u>Step 1:</u> The balanced equation
HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)
It takes 1 mole of NaOH to neutralize 1 mole of the triprotic acid. This is called the reaction stoichiometry.
<u>Step 2:</u> Data given
Mass of the acid = 0.093 grams
volume = 250 mL
titrates with 0.16 M NaOH
adds 6.5 mL NaOH
<u>Step 3: </u>Calculate moles of NaOH
We know the concentration and volume of NaOH needed to neutralize the acid.
By determining the moles of NaOH in that volume in liters (95.9mL=0.0959L), the moles of acid in the original sample can be determined from the reaction stoichiometry.
Moles = Molarity * Volume
Moles = 0.16 M * 0.0065 L
Moles = 0.00104 moles NaOH
<u>Step 4: </u>Calculate moles of the unknown acid:
It takes 1 mole of NaOH to neutralize 1 mole of the triprotic acid. This is called the reaction stoichiometry.
For 0.00104 moles NaOH we have 0.00104 moles of HA
<u>Step 5: </u>Calculate the molar mass of the acid
Molar mass Ha = Mass Ha / moles Ha
Molar mass Ha = 0.093 grams / 0.00104 moles
Molar mass Ha = 89.42 g/mol ≈89 g/mol
The molar mass of the unknown acid is 89 g/mol
