102: Tartaric acid, H2C4H406, has two acidic hydrogens. The acid is often present in wines and precipitates from solution as the wine ages. A solution containing an unknown concentration of the acid is titrated with NaOH. It requires 24.65 ml of 0.2500 MNaOH solution to titrate both acidic protons in 50.00 ml of the tartaric acid solution. Write a balanced net ionic equation for the neutralization reaction, and calculate the molarity of the tartaric acid solution.
Solution 102: Solution: Here, we are going to write the net ionic equation for the given neutralization reaction. Also, we are going to calculate the molarity of the tartaric acid solution. Step1: Tartaric acid reacts with sodium hydroxide to give the following neutralization reaction: H2 4 4 +6NaOH -------> Na C H O + 2H2O4 4 6 2 While writing the ionic equation, we write out all the soluble compounds as ions and eliminate ions common to both the reactants and products. This will give us the net ionic equation. So, the balanced ionic equation equation for the above reaction is: H2 4 4 +6Na + 2OH --------> 2Na + C H O + 2H O 4 4 6- 2 Eliminating the spectator ions from both sides we get the net ionic equation as shown below: - 2- H2 4 4 +6OH --------> C H O + 2H4O4 6 2 Step2: 24.65 Now, Volume of the NaOH solution used = 24.65 mL = 1000 L = 0.02465 L Molarity of the NaOH solution = 0.2500 M Therefore, number of moles of NaOH = Volume of solution in litres X Molarity = 0.02465 L X 0.2500 M = 0.006 mol Step3: From the equation given above, it is sen that, 2 mol of NaOH neutralizes 1 mol of tartaric acid. Therefore, 0.006 mol of NaOH will neutralize ( x 0.006 = 0.003) mol o2 tartaric acid. Step4: Now, number of moles of tartaric acid = 0.003 mol 50 Volume of the tartaric acid solution = 50 mL = 1000 L = 0.050 L Therefore, molarity of the tartaric acid solution = number of moles of tartaric acid / volume of solution in litres. = 0.003 mol / 0.050 L = 0.06 M Thus, the molarity of the tartaric acid solution is 0.06 M.