25.0 mL of a 0.1067 M Solution of nickel (II) sulfate is diluted to 250 mL with water. 10 mL of this solution is taken and diluted to 1.0 L.
Calculate the concentration of [Ni2+] in the final solution in ppm.
0.1067M x (25/250) x (10/1000) = ?
or you can do this in you head.
25 to 250 is 1/10 and 10 to 1000 is 1/100 so the final concn is just 1/1000 x 0.1067 = 0.0001067M
To calculate the concentration of [Ni2+] in the final solution in ppm, we need to follow the given steps:
Step 1: Calculate the moles of nickel (II) sulfate in the original solution.
Moles = concentration × volume in liters
Given concentration = 0.1067 M
Given volume = 25.0 mL = 0.025 L
Moles = 0.1067 M × 0.025 L = 0.002667 moles Ni2+
Step 2: Calculate the moles of nickel (II) sulfate in the 10 mL solution.
Moles = concentration × volume in liters
Given concentration = 0.002667 moles Ni2+
Given volume = 10 mL = 0.01 L
Moles = 0.002667 moles Ni2+ × 0.01 L = 0.00002667 moles Ni2+
Step 3: Calculate the moles of nickel (II) sulfate in the final solution.
Moles = concentration × volume in liters
Given concentration = 0.00002667 moles Ni2+
Given volume = 1.0 L
Moles = 0.00002667 moles Ni2+ × 1.0 L = 0.00002667 moles Ni2+
Step 4: Calculate the concentration of [Ni2+] in the final solution in ppm.
ppm = (moles of solute ÷ volume of solution) × 10^6
Given moles = 0.00002667 moles Ni2+
Given volume = 1.0 L = 1000 mL
ppm = (0.00002667 moles ÷ 1000 mL) × 10^6 = 26.67 ppm
Therefore, the concentration of [Ni2+] in the final solution is 26.67 ppm.
To calculate the concentration of [Ni2+] in the final solution in parts per million (ppm), we need to follow a step-by-step approach:
Step 1: Calculate the moles of Ni2+ in the original solution.
First, we need to calculate the moles of Ni2+ in the original solution using the volume and molarity of the solution.
Moles of Ni2+ = Volume (L) × Molarity
= 0.025 L × 0.1067 mol/L
= 0.0026675 mol
Step 2: Calculate the moles of Ni2+ in the final solution.
Now, we need to calculate the moles of Ni2+ in the final solution by taking 10 mL (0.01 L) of the original solution and diluting it to 1.0 L.
Moles of Ni2+ in final solution = Moles of Ni2+ in original solution × (Volume of final solution / Volume of original solution)
= 0.0026675 mol × (1.0 L / 0.01 L)
= 0.26675 mol
Step 3: Calculate the mass of Ni2+ in the final solution.
To calculate the mass of Ni2+ in the final solution, we multiply the moles of Ni2+ in the final solution by the molar mass of Ni2+.
Molar mass of Ni2+ = Atomic mass of Ni × 2
= 58.69 g/mol × 2
= 117.38 g/mol
Mass of Ni2+ in final solution = Moles of Ni2+ in final solution × Molar mass of Ni2+
= 0.26675 mol × 117.38 g/mol
= 31.34 g
Step 4: Calculate the concentration of Ni2+ in ppm.
Finally, we calculate the concentration of Ni2+ in the final solution in parts per million (ppm) using the mass of Ni2+ in the final solution and the volume of the final solution.
Concentration of Ni2+ in ppm = (Mass of Ni2+ / Volume of final solution) × 10^6
= (31.34 g / 1.0 L) × 10^6
= 3.134 × 10^7 ppm
Therefore, the concentration of [Ni2+] in the final solution is 3.134 × 10^7 ppm.