15ml of 0.309M Na2SO4 and 35.6ml of 0.200M KCL are mixed determine the following concentration (Na+),(SO4-2),(CL-1)
To determine the concentrations of Na+, SO4-2, and Cl- in the solution, we need to use the concept of stoichiometry and the information given about the volumes and concentrations of the two solutes.
Step 1: Calculate the number of moles of Na2SO4 and KCl.
Moles of Na2SO4 = Volume (in L) × Concentration (in mol/L)
Moles of Na2SO4 = 15 ml × (1 L/1000 ml) × 0.309 mol/L
Moles of Na2SO4 = 0.004635 mol
Moles of KCl = Volume (in L) × Concentration (in mol/L)
Moles of KCl = 35.6 ml × (1 L/1000 ml) × 0.200 mol/L
Moles of KCl = 0.00712 mol
Step 2: Determine the number of moles of Na+, SO4-2, and Cl- ions.
In Na2SO4, there are 2 moles of Na+ ions and 1 mole of SO4-2 ions.
In KCl, there is 1 mole of K+ ions and 1 mole of Cl- ions.
Number of moles of Na+ ions = 2 × Moles of Na2SO4
Number of moles of Na+ ions = 2 × 0.004635 mol
Number of moles of Na+ ions = 0.00927 mol
Number of moles of SO4-2 ions = 1 × Moles of Na2SO4
Number of moles of SO4-2 ions = 1 × 0.004635 mol
Number of moles of SO4-2 ions = 0.004635 mol
Number of moles of Cl- ions = 1 × Moles of KCl
Number of moles of Cl- ions = 1 × 0.00712 mol
Number of moles of Cl- ions = 0.00712 mol
Step 3: Calculate the concentrations of Na+, SO4-2, and Cl- ions.
Concentration (in mol/L) = Number of moles / Volume (in L)
Concentration of Na+ ions = 0.00927 mol / (15 ml × 1 L/1000 ml)
Concentration of Na+ ions = 0.00927 mol / 0.015 L
Concentration of Na+ ions = 0.618 M (rounded to three significant figures)
Concentration of SO4-2 ions = 0.004635 mol / (15 ml × 1 L/1000 ml)
Concentration of SO4-2 ions = 0.004635 mol / 0.015 L
Concentration of SO4-2 ions = 0.309 M (rounded to three significant figures)
Concentration of Cl- ions = 0.00712 mol / (35.6 ml × 1 L/1000 ml)
Concentration of Cl- ions = 0.00712 mol / 0.0356 L
Concentration of Cl- ions = 0.200 M
So, the concentrations of Na+, SO4-2, and Cl- ions in the solution are approximately 0.618 M, 0.309 M, and 0.200 M, respectively.