If 32.6 mL of 0.240 M oxalic acid are required to titrate 48.9 mL of unknown sodium hydroxide: How many moles of acid are used in the titration? How many moles of base are used in the titration? What is the molarity of the unknown NaOh base?
Please help I do not know where to start.
You start by writing an equation.
2NaOH + H2C2O4 ==> Na2C2O4 + 2H2O
mols H2C2O4 = M x L = ?
Using the coefficients in the balanced equation, convert mols H2C2O4 to mols NaOH.
Then M NaOH = mols NaOH/L NaOH
To solve this problem, we can use the concept of stoichiometry and the equation for the reaction between oxalic acid (H2C2O4) and sodium hydroxide (NaOH):
H2C2O4 + 2NaOH -> Na2C2O4 + 2H2O
1. Calculate the number of moles of oxalic acid used in the titration:
Number of moles = concentration (M) * volume (L)
Given: concentration = 0.240 M, volume = 32.6 mL (convert to L: 32.6 mL / 1000 mL/L)
Moles of oxalic acid = 0.240 M * 32.6 mL / 1000 = 0.007824 moles
2. Since the balanced equation shows a 1:2 molar ratio between oxalic acid and sodium hydroxide, we can conclude that the number of moles of sodium hydroxide used in the titration is twice the number of moles of oxalic acid. Thus:
Moles of sodium hydroxide = 2 * 0.007824 = 0.015648 moles
3. To find the molarity of the unknown sodium hydroxide, we can use this formula:
Molarity (M) = moles / volume (L)
Given: volume = 48.9 mL (convert to L: 48.9 mL / 1000 mL/L)
Molarity of sodium hydroxide = 0.015648 moles / 48.9 mL / 1000 = 0.320 M
So, to recap:
- The number of moles of oxalic acid used in the titration is 0.007824 moles.
- The number of moles of sodium hydroxide used in the titration is 0.015648 moles.
- The molarity of the unknown sodium hydroxide solution is 0.320 M.
To solve this problem, we need to use the concept of stoichiometry and the equation of the reaction between oxalic acid and sodium hydroxide.
The equation for the reaction between oxalic acid (H2C2O4) and sodium hydroxide (NaOH) is as follows:
2H2C2O4 + 2NaOH → Na2C2O4 + 2H2O
From the balanced equation, we can see that for every 2 moles of oxalic acid, we need 2 moles of sodium hydroxide to react.
Now let's calculate the number of moles of oxalic acid used in the titration:
First, we convert the volume of oxalic acid (32.6 mL) to liters by dividing it by 1000:
32.6 mL ÷ 1000 = 0.0326 L
Next, we use the given molarity of oxalic acid (0.240 M) to calculate the number of moles of oxalic acid:
Number of moles = Molarity × Volume
Number of moles = 0.240 M × 0.0326 L
Calculating this gives us:
Number of moles of oxalic acid = 0.00782 moles (rounded to 4 decimal places)
Since the stoichiometry of the reaction is 2:2, the number of moles of sodium hydroxide used in the titration will be the same as the number of moles of oxalic acid used.
Therefore, the number of moles of base used in the titration is also 0.00782 moles.
To find the molarity of the unknown sodium hydroxide solution, we need to use the volume and the number of moles of the base.
We are given the volume of the unknown sodium hydroxide solution as 48.9 mL. Again, we convert it to liters:
48.9 mL ÷ 1000 = 0.0489 L
Finally, we use the relationship between moles, molarity, and volume to find the molarity of the base:
Molarity = Number of moles ÷ Volume
Molarity = 0.00782 moles ÷ 0.0489 L
Calculating this gives us:
Molarity of the unknown NaOH base = 0.16 M (rounded to 2 decimal places)
Therefore, the molarity of the unknown NaOH base is 0.16 M.