A given solution contained potassium chloride, KCl, and hydrochloric acid. 25 cm3 of it needed 24.80 cm3 of 0.0986 M sodium hydroxide solution to neutralize the acid, and the neutral solution then required 23.55 cm3 of 0.2 M silver nitrate, AgNO3, solution to precipitate all of the chloride as silver chloride, AgCl. Calculate the concentration in g dm-3 of potassium chloride and hydrochloric acid in the solution.
mols HCl = M NaOH x L NaOH - ?
mols KCl+HCl = M AgNO3 x L AgNO3 = ?
mols KCl = (mols KCl+HCl)-mols HCl(from NaOH data)
g HCl = mols HCl x molar mass HCl = ?g HCl in the 25 cc.
g KCl = mols KCl x molar mass KCl = ?g KCl in th 25 cc.
Convert g in 25 cc to g/dm^3
To calculate the concentration in g dm-3 of potassium chloride (KCl) and hydrochloric acid (HCl) in the given solution, we can use the information provided about the volume and concentration of the solutions involved, as well as the stoichiometry of the reactions.
1. Start by identifying the balanced chemical equations for the reactions involved:
a) Reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH):
HCl + NaOH -> NaCl + H2O
b) Reaction between potassium chloride (KCl) and silver nitrate (AgNO3):
KCl + AgNO3 -> AgCl + KNO3
2. Calculate the number of moles of sodium hydroxide (NaOH) used:
Given volume of NaOH solution = 24.80 cm3 = 24.80 / 1000 dm3 = 0.02480 dm3
Given concentration of NaOH = 0.0986 M
Number of moles of NaOH = concentration x volume = 0.0986 M x 0.02480 dm3 = 0.00244 moles
Since the reaction between HCl and NaOH is 1:1, this indicates that the number of moles of hydrochloric acid (HCl) in the original solution is also 0.00244 moles.
3. Calculate the number of moles of silver nitrate (AgNO3) used:
Given volume of AgNO3 solution = 23.55 cm3 = 23.55 / 1000 dm3 = 0.02355 dm3
Given concentration of AgNO3 = 0.2 M
Number of moles of AgNO3 = concentration x volume = 0.2 M x 0.02355 dm3 = 0.00471 moles
Since the reaction between KCl and AgNO3 is 1:1, this indicates that the number of moles of KCl in the original solution is also 0.00471 moles.
4. Calculate the concentrations in g dm-3 of KCl and HCl:
a) Concentration of KCl:
Mass of KCl = number of moles x molar mass = 0.00471 moles x (39.1 g/mol + 35.5 g/mol) = 0.349 g
Volume of the solution = 25 cm3 = 25 / 1000 dm3 = 0.025 dm3
Concentration of KCl = mass/volume = 0.349 g / 0.025 dm3 = 13.96 g dm-3
b) Concentration of HCl:
Mass of HCl = number of moles x molar mass = 0.00244 moles x (1 g/mol + 35.5 g/mol) = 0.087 g
Volume of the solution = 25 cm3 = 25 / 1000 dm3 = 0.025 dm3
Concentration of HCl = mass/volume = 0.087 g / 0.025 dm3 = 3.48 g dm-3
Therefore, the concentration of potassium chloride in the solution is 13.96 g dm-3, and the concentration of hydrochloric acid is 3.48 g dm-3.