The quantity of NaOH required to reach the third equivalence point is 20mL.
Using the titration formula,
CaVa/CbVb = Na/Nb
Where,
Ca = concentration of citric acid (0.200 M)
Cb = concentration of NaOH (0.750 M)
Va = Volume of citric acid (25.0 mL)
Vb = volume of NaOH (x mL)
Na = number of reacting mole of citric acid (3)
Nb = number of reacting mole of NaOH (1)
Therefore Vb ( x mL) =CaVaNb/CbNa
= 0.2× 25×3/0.75 ×1
= 15/0.75
Vb ( x mL) = 20 mL
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Answer:
1.09 L
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the volume in liters of a 0.360 mol/L barium acetate solution that contains 100 g of barium acetate. Be sure your answer has the correct number of significant digits.</em>
<em />
The molar mass of barium acetate is 255.43 g/mol. The moles corresponding to 100 grams are:
100 g × (1 mol/255.43 g) = 0.391 mol
0.391 moles of barium acetate are contained in an unknown volume of a 0.360 mol/L barium acetate solution. The volume is:
0.391 mol × (1 L/0.360 mol) = 1.09 L
Complete Question:
A chemist prepares a solution of iron chloride by measuring out 0.10 g of FeCl2 into a 50. mL volumetric flask and filling to the mark with distilled water. Calculate the molarity of anions in the chemist's solution.
Answer:
[Fe+] = 0.0156 M
[Cl-] = 0.0316 M
Explanation:
The molar mass of iron chloride is 126.75 g/mol, thus, the number of moles presented in 0.10 g of it is:
n = mass/molar mass
n = 0.10/126.75
n = 7.89x10⁻⁴ mol
In a solution, it will dissociate to form:
FeCl2 -> Fe+ + 2Cl-
So, the stoichiometry is 1:1:2, and the number of moles of the ions formed are:
nFe+ = 7.89x10⁻⁴ mol
nCl- = 2*7.89x10⁻⁴ = 1.58x10⁻³ mol
The molarity is the number of moles divided by the solution volume, in L (50.0 mL = 0.05 L):
[Fe+] = 7.89x10⁻⁴/0.05 = 0.0156 M
[Cl-] = 1.58x10⁻³/0.05 = 0.0316 M
