25ml of sulfuric acid from a car battery was accurately diluted to 500ml. 25ml of the diluted solution was titrated with 0.206 mol/L sodium hydroxide. It required 38.8 ml. Calculate the Molarity of the original battery acid and it's concentration in grams per litre.
Answer=3.20M
I'm not sure how to get the answer, I'm pretty sure it relates to c1v1=c2v2.
Thanks.
sulfuric acid is di-protic (two H+)
1/2 * (500 / 25) * .206 * 38.8 / 25
Answer of the question please.
Please give me the solution of the question that you are hiding.....
You're correct, the problem can be solved using the formula C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
To start, let's calculate the initial concentration of the sulfuric acid.
C1V1 = C2V2
C1 * 25 ml = 0.206 mol/L * 38.8 ml
Rearranging the equation to solve for C1:
C1 = (0.206 mol/L * 38.8 ml) / 25 ml
C1 ≈ 0.3208 mol/L
So, the initial concentration of the sulfuric acid is approximately 0.3208 mol/L.
Next, let's calculate the concentration in grams per liter (g/L).
To do this, we need to know the molar mass of sulfuric acid, which is H2SO4.
H (hydrogen) = 1.00784 g/mol
S (sulfur) = 32.06 g/mol
O (oxygen) = 16.00 g/mol
H2SO4 has 2 hydrogen atoms, 1 sulfur atom, and 4 oxygen atoms. So, we can calculate the molar mass of sulfuric acid as follows:
(2 * 1.00784 g/mol) + (1 * 32.06 g/mol) + (4 * 16.00 g/mol)
= 98.08 g/mol
Now, let's calculate the concentration in grams per liter.
Concentration in g/L = (C1 * molar mass of sulfuric acid)
= (0.3208 mol/L * 98.08 g/mol)
≈ 31.70 g/L
Therefore, the Molarity of the original battery acid is 3.20 M and its concentration in grams per liter is approximately 31.70 g/L.