How many grams of citric acid is needed to titrate 20ml NaOH 0.05M
their mole ratio is 1:3 respectively
To determine the number of grams of citric acid needed to titrate 20 mL of NaOH solution with a concentration of 0.05 M, we first need to determine the mole ratio between citric acid and NaOH.
From the given information, the mole ratio is 1:3, which means that one mole of citric acid reacts with three moles of NaOH.
Next, we calculate the number of moles of NaOH in 20 mL of a 0.05 M solution. To do this, we use the formula:
moles = concentration (M) x volume (L)
Converting the volume in mL to liters:
volume (L) = 20 mL / 1000 = 0.02 L
Now, we can calculate the number of moles of NaOH:
moles of NaOH = 0.05 M x 0.02 L = 0.001 mol
Since the mole ratio between citric acid and NaOH is 1:3, we know that 1 mole of citric acid is required to react with 3 moles of NaOH.
Therefore, to find the number of moles of citric acid required, we divide the moles of NaOH by the mole ratio:
moles of citric acid = 0.001 mol NaOH / 3 = 0.00033 mol
Finally, the last step is to convert moles of citric acid to grams. We need to know the molar mass of citric acid to do this calculation. The molar mass of citric acid is approximately 192.13 g/mol.
mass of citric acid = moles of citric acid x molar mass of citric acid
mass of citric acid = 0.00033 mol x 192.13 g/mol ≈ 0.06316 g
Therefore, approximately 0.06316 grams of citric acid is needed to titrate 20 mL of NaOH solution with a concentration of 0.05 M, given a mole ratio of 1:3.