Tartaric acid has two acidic (ionizable) hydrogens. The acid is often present in wines and precipitates from solution as wine ages. A solution containing an unknown concentration of the acid is titrated with NaOH. It requires 22.62 mL of 0.2000 M NaOH solution of titrate both acidic protons in 40.00 mL of tartaric acid solution. Write the balanced molecular and net ionic equations for the neutralization reaction and calculate the molarity of the tartaric acid solution.
To start, let's write the balanced molecular and net ionic equations for the neutralization reaction between tartaric acid (H2C4H4O6) and NaOH:
Balanced Molecular Equation:
H2C4H4O6 + 2NaOH -> Na2C4H4O6 + 2H2O
Balanced Net Ionic Equation:
H2C4H4O6 + 2OH- -> C4H4O6 2- + 2H2O
In the balanced net ionic equation, we only include the species involved in the reaction and ignore spectator ions (ions that appear on both sides of the equation and do not participate in the reaction).
Now let's calculate the molarity of the tartaric acid solution using the titration data given:
- 22.62 mL of 0.2000 M NaOH solution is required to neutralize both acidic protons in 40.00 mL of tartaric acid solution.
First, let's determine the number of moles of NaOH used in the titration:
moles of NaOH = volume of NaOH solution (L) x molarity of NaOH (mol/L)
moles of NaOH = 0.02262 L x 0.2000 mol/L
moles of NaOH = 0.004524 mol
Since the balanced equation from above shows that the molar ratio between NaOH and tartaric acid is 2:1, we can determine the number of moles of tartaric acid in the solution:
moles of tartaric acid = 0.004524 mol x 1/2
moles of tartaric acid = 0.002262 mol
Next, we need to calculate the molarity (M) of the tartaric acid solution:
Molarity (M) = moles of tartaric acid (mol) / volume of tartaric acid solution (L)
Molarity of tartaric acid = 0.002262 mol / 0.04000 L
Molarity of tartaric acid = 0.0566 M
Therefore, the molarity of the tartaric acid solution is 0.0566 M.