1.25 grams of an acidnis completely neutralised by 25ml of a 0.25M Ba(OH)2 SOLUTION.which of the following statements are correct.
a)if the acid is dibasic its molar mass would be 200
b)if it is dibasic /0.5g of the acid would neutralise completely 12.5ml of 0.40N NaOH.
C)volume of base used is independent of strength of the acid
To determine which of the statements are correct, let's analyze the given information step-by-step:
1. First, we need to determine the number of moles of Ba(OH)2 used for neutralization:
Moles of Ba(OH)2 = Molarity x Volume
Moles of Ba(OH)2 = 0.25 mol/L x 0.025 L
Moles of Ba(OH)2 = 0.00625 mol
2. Since the stoichiometric ratio between Ba(OH)2 and the acid is 1:2 (assuming the acid is dibasic), the number of moles of acid is twice that of Ba(OH)2:
Moles of acid = 2 x Moles of Ba(OH)2
Moles of acid = 2 x 0.00625 mol
Moles of acid = 0.0125 mol
3. To find the molar mass of the acid, divide its mass by the number of moles:
Molar mass = Mass / Moles
Molar mass = 1.25 g / 0.0125 mol
Molar mass = 100 g/mol
Statement a) "If the acid is dibasic, its molar mass would be 200" is incorrect. The molar mass of the acid is actually 100 g/mol based on the calculations.
4. Next, let's evaluate statement b) "If it is dibasic, 0.5 g of the acid would neutralize completely 12.5 ml of 0.4N NaOH."
To check this statement, we need to find the number of moles of NaOH used for neutralization:
Moles of NaOH = Molarity x Volume
Moles of NaOH = 0.4 mol/L x 0.0125 L
Moles of NaOH = 0.005 mol
Since the stoichiometric ratio between the acid and NaOH is 1:2 (assuming the acid is dibasic), the number of moles of acid is also twice that of NaOH:
Moles of acid = 2 x Moles of NaOH
Moles of acid = 2 x 0.005 mol
Moles of acid = 0.01 mol
To find the mass of the acid required, we can use the molar mass:
Mass = Moles x Molar mass
Mass = 0.01 mol x 100 g/mol
Mass = 1 g
The statement mentions that 0.5 g of the acid would neutralize completely 12.5 ml of 0.4N NaOH, but based on the calculations, 1 g of the acid is required for neutralization. Therefore, statement b) is incorrect.
Lastly, let's evaluate statement c) "Volume of base used is independent of the strength of the acid."
From the given information, it is clear that the volume of Ba(OH)2 solution used is 25 ml, regardless of the strength of the acid. Therefore, statement c) is correct.
In summary:
a) Incorrect. The molar mass of the acid is 100 g/mol, not 200 g/mol.
b) Incorrect. 0.5 g of the acid is not sufficient to neutralize 12.5 ml of 0.4N NaOH; instead, 1 g of the acid is required.
c) Correct. The volume of the base used is independent of the strength of the acid.
To determine which of the statements are correct, let's analyze the given information:
1.25 grams of an acid are completely neutralized by 25 ml of a 0.25M Ba(OH)2 solution.
a) Statement a) If the acid is dibasic, its molar mass would be 200.
To determine if the molar mass of the acid is 200, we can use the given information and the concept of moles. Since we know the molarity (0.25 M) and volume (25 ml) of the Ba(OH)2 solution used to neutralize the acid, we can calculate the number of moles of Ba(OH)2.
Moles of Ba(OH)2 = Molarity x Volume (in L)
Moles of Ba(OH)2 = 0.25 M x 0.025 L = 0.00625 moles
According to the neutralization reaction, each mole of Ba(OH)2 reacts with two moles of acid, since it is stated that the acid is dibasic. Therefore, the number of moles of acid would be twice the number of moles of Ba(OH)2.
Moles of acid = 2 x Moles of Ba(OH)2 = 2 x 0.00625 moles = 0.0125 moles
To calculate the molar mass of the acid, we can use the equation: Molar mass = Mass / Moles.
Molar mass = 1.25 g / 0.0125 moles = 100 g/mol
Therefore, statement a) is not correct, as the molar mass of the acid would be 100 g/mol, not 200 g/mol.
Next, let's consider statement b):
b) If it is dibasic, 0.5g of the acid would neutralize completely 12.5ml of 0.40N NaOH.
To determine if this statement is correct, we can use the concept of molarity and neutralization reactions.
0.25 M Ba(OH)2 solution is equivalent to 0.25 moles of Ba(OH)2 in 1 liter of solution.
0.40 N NaOH solution is equivalent to 0.40 moles of NaOH in 1 liter of solution.
Since NaOH and Ba(OH)2 react in a 1:2 ratio, 0.40 moles of NaOH can neutralize 0.20 moles of Ba(OH)2.
Now, let's determine how many moles of acid can be neutralized by these 0.20 moles of Ba(OH)2:
Since the acid is dibasic, 1 mole of acid will react with 2 moles of Ba(OH)2.
Therefore, 0.20 moles of Ba(OH)2 will be able to neutralize 0.10 moles of acid.
Now, we can calculate how many grams of the acid can be neutralized by these 0.10 moles:
Mass of the acid = Moles x Molar mass
Mass of the acid = 0.10 moles x Molar mass
0.5 g = 0.10 moles x Molar mass
Molar mass = 0.5 g / 0.10 moles = 5 g/mol
Therefore, statement b) is not correct, as the molar mass of the acid would be 5 g/mol, not 200 g/mol.
Finally, let's consider statement c):
c) The volume of base used is independent of the strength of the acid.
This statement is correct. From the given information, we can see that the volume of the Ba(OH)2 solution used for neutralization is 25 ml, regardless of the strength of the acid. The volume of the base used for neutralization is determined by the stoichiometry of the reaction rather than the strength of the acid.
Therefore, the correct statement is c) - the volume of base used is independent of the strength of the acid.
The only way I know to do this is to work each answer as a separate problem.
a. H2A + Ba(OH)2 ==> BaA + 2H2O
mols Ba(OH)2 = M x L = ?
mols = grams/molar mass. You have mols from above and grams from the problem, solve for molar mass. If you get 200 for molar mass then a must be right.
b. mols H2A = grams/molar mass. Use 0.5 g as in the problem and use molar mass found in part a. Then
H2A + 2NaOH ==> Na2A + 2H2O
Convert mols H2A to mols NaOH needed and compare with mols generated by 12.5 mL of 0.40M NaOH. YOu do that with mols = M x L = ? This answer will tell you if b is correct or not.
c. C is a dumb answer; you know it can't be right.