A solution of dibasic acid H2X contains 12.60gr of acid per litre, 10cm^3 of this acid solution required 20cm^3 of 0.1M potassium hydroxide solution for complete neutralization using phenolphthalein
Find relative molecular mass of of acid and deduce the atomic mass of X in formula H2X
H2X + 2KOH ==> K2X + 2H2O
mols = M x L = 0.1 M x 0.020 L = 0.002 mols KOH
Looking at the equation you know 1 mol H2X = 2 mols kOH; therefore, 2 moles KOH = 1 mol H2X = 0.001 mol H2X in 10 cc of the solution or 0.001 x 1000/10 = 0.1 mol H2X/L and that = 12.6 g H2X so 12.6 g H2X = 0.1 mol.
moles = grams/molar mass or
molar mass = grams/mols = 12.6/0.1 = 126 = molar mass H2X.
From that we know H is 2 for the two so X must be 124. I don't know of any element with atomic mass 124. You might hazard a guess at Te with atomic mass 127.6 which would make H2X = about 130.The only problem I have with that is that the accuracy of a titration like that is FAR BETTER than that. So I think there must be something wrong with the problem or the post has a typo. I'm a shaggy analytical chemist if you give me a sample of H2X with a molar mass of 130 and I can't do better than 126 when I titrate it.
so can you know that hiden element in that dibasic acid
To find the relative molecular mass of the acid H2X, we can use the information given about the reaction with potassium hydroxide.
1. Convert the volume of potassium hydroxide solution used to moles:
Molarity (M) = moles/volume (L)
0.1 M = moles / 0.020 L
Moles of KOH = 0.1 * 0.020 = 0.002
2. According to the neutralization reaction, the ratio between the acid and the base is 1:1. Therefore, the moles of H2X is also 0.002.
3. Calculate the mass of the acid in 10 cm^3 (0.01 L) of the solution:
Mass of H2X = (12.60 g/L) * 0.01 L = 0.126 g
4. Find the molar mass of H2X:
Molar mass = mass/moles
Molar mass = 0.126 g / 0.002 = 63 g/mol
5. Deduce the atomic mass of X in the formula H2X:
The molar mass of H2X is 63 g/mol. Since H2 contributes 2 g/mol to the molar mass, the atomic mass of X can be found by subtracting the molar mass of H2 from the molar mass of H2X:
Atomic mass of X = Molar mass of H2X - Molar mass of H2
Atomic mass of X = 63 g/mol - 2 g/mol = 61 g/mol
Therefore, the relative molecular mass of the dibasic acid H2X is 63 g/mol, and the atomic mass of X in the formula H2X is 61 g/mol.
To find the relative molecular mass of the acid and deduce the atomic mass of X in the formula H2X, we can use the concept of stoichiometry.
1. First, let's calculate the number of moles of potassium hydroxide (KOH) used in the neutralization reaction. We know that the volume of the KOH solution is 20 cm^3 and its concentration is 0.1 M (moles per liter). Therefore:
Moles of KOH = Volume (in liters) x Concentration
= 20 cm^3 x (1 liter/1000 cm^3) x 0.1 M
= 0.002 moles
2. Since the reaction is balanced in a 1:1 ratio between the acid and the base, the moles of the acid will also be 0.002.
3. We are given that the acid solution contains 12.60 g of acid per liter. So, the mass of the acid used in the reaction (in grams) can be calculated using the volume of the acid solution used, which is 10 cm^3 (1 cm^3 = 1 mL).
Mass of acid = Volume (in liters) x Concentration
= 10 cm^3 x (1 liter/1000 cm^3) x 12.60 g/L
= 0.126 g
4. Now, we can calculate the relative molecular mass of the acid using the equation:
Relative molecular mass = Mass of acid (in grams) / Moles of acid
Relative molecular mass = 0.126 g / 0.002 mol
= 63 g/mol
5. Since the formula of the acid is H2X and we know that the relative molecular mass is 63 g/mol, we can deduce the atomic mass of X as follows:
The relative molecular mass of the acid (H2X) = Atomic mass of H (2 atoms) + Atomic mass of X (1 atom)
63 g/mol = (2 x Atomic mass of H) + Atomic mass of X
We can use the known atomic mass of hydrogen (1 g/mol) to solve for X:
63 g/mol = (2 x 1 g/mol) + Atomic mass of X
63 g/mol = 2 g/mol + Atomic mass of X
Atomic mass of X = 63 g/mol - 2 g/mol
Atomic mass of X = 61 g/mol
Therefore, the atomic mass of X in the formula H2X is 61 g/mol.