A 10 mL sample of tartaric acid (a diprotic acid) is titrated with 20 mL of 1.0 M NaOH. What is the molarity of the acid?
2(M)(.010)=.02*1
solve for M
Bakrii
M1
To determine the molarity of the tartaric acid, we need to use the concept of stoichiometry and the balanced equation of the reaction between the tartaric acid (HA) and sodium hydroxide (NaOH):
HA + NaOH -> NaA + H2O
From the balanced equation, we can see that the mole ratio between the tartaric acid (HA) and the NaOH is 1:1. This means that one mole of tartaric acid reacts with one mole of NaOH.
Given the volume and molarity of the NaOH solution, we can calculate the number of moles of NaOH used in the titration:
moles of NaOH = volume of NaOH solution (in liters) * molarity of NaOH
= 0.020 L * 1.0 mol/L
= 0.020 mol
Since the mole ratio between the tartaric acid and NaOH is 1:1, the number of moles of tartaric acid in the sample is also 0.020 mol.
Now, we need to find the molarity of the tartaric acid. The molarity (M) is defined as moles of solute divided by the volume of the solution in liters:
Molarity of tartaric acid = moles of tartaric acid / volume of tartaric acid (in liters)
In the given problem, we have a 10 mL sample of tartaric acid, which is equivalent to 0.010 L:
Molarity of tartaric acid = 0.020 mol / 0.010 L
= 2.0 M
Therefore, the molarity of the tartaric acid is 2.0 M.