How much cm3 solution of concentrated ammonia with 25% density of 0.91g/cm3 should be added to 7 g NH4Cl so we can get a solution of 100cm3 with pH 10 (Kb = 1.8 x 10-5)
To find the amount of concentrated ammonia solution needed, we first need to calculate the number of moles of NH4Cl and NH3 involved in the reaction.
1. Calculate the number of moles of NH4Cl:
We are given the mass of NH4Cl (7 g), and we can determine the molar mass of NH4Cl to convert grams to moles.
The molar mass of NH4Cl = 14.01 g/mol (atomic weight of nitrogen) + 4 * (1.01 g/mol) (4 hydrogen atoms) + 35.45 g/mol (atomic weight of chlorine)
So, the molar mass of NH4Cl = 53.49 g/mol.
Now, we can calculate the number of moles of NH4Cl using the formula:
Moles of NH4Cl = Mass of NH4Cl / Molar mass of NH4Cl = 7 g / 53.49 g/mol = 0.131 mol.
2. Calculate the number of moles of NH3:
The reaction between NH4Cl and NH3 is as follows:
NH4Cl + NH3 -> NH4OH + HCl
We can see that one mole of NH4Cl reacts with one mole of NH3 in the reaction. Therefore, the number of moles of NH3 is also 0.131 mol.
3. Calculate the number of moles of NH3 from the volume and concentration:
We are given that the final solution volume is 100 cm^3.
To convert the volume to dm^3, we need to divide by 1000:
Volume = 100 cm^3 / 1000 cm^3/dm^3 = 0.1 dm^3.
The concentration of NH3 is given as 25%, which means there is 25 g of NH3 in 100 g of solution.
We can calculate the mass of NH3 required using the formula:
Mass of NH3 = (0.25 * Mass of solution) = (0.25 * 0.91 g/cm^3 * 0.1 dm^3) = 0.02275 g
Now, we can calculate the number of moles of NH3 using the molar mass of NH3 (17.03 g/mol):
Moles of NH3 = Mass of NH3 / Molar mass of NH3 = 0.02275 g / 17.03 g/mol = 0.00134 mol.
4. Determine the excess amount of NH3:
The initial number of moles of NH3 (0.131 mol) is greater than the required number of moles of NH3 (0.00134 mol).
The excess amount of NH3 needed can be calculated by subtracting the required amount from the initial amount:
Excess amount of NH3 = Initial number of moles - Required number of moles = 0.131 mol - 0.00134 mol = 0.12966 mol.
Therefore, we need to add a solution with an excess amount of 0.12966 mol (or 0.131 mol, rounded to three significant figures) of NH3 to obtain a final solution volume of 100 cm^3 with a pH of 10.