In a titration experiment, 31. 4 ml of 1. 120 m hcooh is neutralized by 16. 3 ml of ba(oh)2. what is the concentration of the ba
1 answer:
In a titration experiment, 31. 4 ml of 1. 120 M HCOOH is neutralized by 16. 3 ml of . So, the concentration of the
solution is 23.16 M
Calculation,
The formula for the dilution of a solution is given as:
=
Where M is molarity and V is the volume of the solution in liters ( L ).
Given data,
= 120 M
=31. 4 ml
= ?
= 16. 3 ml
120 M ×31. 4 ml = × 16. 3 ml
= 120 M ×31. 4 ml / 16. 3 ml =23.16 M
In a titration experiment, the concentration of the solution is 23.16 M
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Answer:
0.558mole of SO₃
Explanation:
Given parameters:
Molar mass of SO₃ = 80.0632g/mol
Mass of S = 17.9g
Molar mass of S = 32.065g/mol
Number of moles of O₂ = 0.157mole
Molar mass of O₂ = 31.9988g/mol
Unknown:
Maximum amount of SO₃
Solution
We need to write the proper reaction equation.
2S + 3O₂ → 2SO₃
We should bear in mind that the extent of this reaction relies on the reactant that is in short supply i.e limiting reagent. Here the limiting reagent is the Sulfur, S. The oxygen gas would be in excess since it is readily availbale.
So we simply compare the molar relationship between sulfur and product formed to solve the problem:
First, find the number of moles of Sulfur, S:
Number of moles of S =
Number of moles of S = = 0.558mole
Now to find the maximum amount of SO₃ formed, compare the moles of reactant to the product:
2 mole of Sulfur produced 2 mole of SO₃
Therefore; 0.558mole of sulfur will produce 0.558mole of SO₃
<u>Answer:</u> The pH of the buffer is 5.25
<u>Explanation:</u>
Let the volume of buffer solution be V
We know that:
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
We are given:
= negative logarithm of acid dissociation constant of weak acid = 4.90
pH = ?
Putting values in above equation, we get:
Hence, the pH of the buffer is 5.25