How to solve titration problem for HC2H3O2 + NaOH= NaC2H3O2 + H2O
The M of NaOH= 0.427
The volume of NaOH= 17.30mL
10.00mL of acid
The second titration calculation is:
The M of NaOH= 0.427
The volume of NaOH=34.60mL
20.00mL of acid
mols NaOH = M x L = ?
mols HC2H3O2 = the same (note the balanced equation and 1 mol NaOH reacts with 1 mol HC2H3O2 which means 1:1).
M HC2H3O2 = mols/L . YOu know mols and L.
To solve these titration problems, we'll use the concept of stoichiometry and the balanced chemical equation. The balanced equation for the reaction between HC2H3O2 (acetic acid) and NaOH (sodium hydroxide) is:
HC2H3O2 + NaOH → NaC2H3O2 + H2O
The coefficients in the balanced equation show the ratio of reactants and products. In this case, the ratio is 1:1 for HC2H3O2 and NaOH, meaning that one mole of HC2H3O2 reacts with one mole of NaOH to produce one mole of NaC2H3O2 and one mole of H2O.
Here are the steps to solve the titration problem:
Step 1: Find the number of moles of NaOH used in each titration.
To find the number of moles, we'll use the formula:
moles = concentration (M) × volume (L)
For the first titration, where the volume of NaOH is 17.30 mL and the concentration of NaOH is 0.427 M:
moles of NaOH = 0.427 M × (17.30 mL / 1000 mL/L) = 0.00735 mol
For the second titration, where the volume of NaOH is 34.60 mL and the concentration of NaOH is 0.427 M:
moles of NaOH = 0.427 M × (34.60 mL / 1000 mL/L) = 0.0148 mol
Step 2: Use the stoichiometry from the balanced equation to find the number of moles of HC2H3O2 used.
Since the ratio of HC2H3O2 to NaOH is 1:1, the number of moles of HC2H3O2 used is equal to the number of moles of NaOH used in each titration.
For the first titration: 0.00735 mol of HC2H3O2
For the second titration: 0.0148 mol of HC2H3O2
Step 3: Calculate the molarity of HC2H3O2 in the solution.
To find the molarity, we'll use the formula:
Molarity = moles of solute / volume of solution (in liters)
For the first titration, where the volume of HC2H3O2 is 10.00 mL:
Molarity of HC2H3O2 = 0.00735 mol / (10.00 mL / 1000 mL/L) = 0.735 M
For the second titration, where the volume of HC2H3O2 is 20.00 mL:
Molarity of HC2H3O2 = 0.0148 mol / (20.00 mL / 1000 mL/L) = 0.740 M
So, the molarity of HC2H3O2 in the solutions for the first and second titrations is 0.735 M and 0.740 M, respectively.