5.000ml of 0.200M Fe(NO3)3 combined with 4.000ml of 0.00200M NaSCN in a 50.00ml volumetric flask filled with water. What is the diluted concentration for the SCN- ion, after the solution is made

Question

5.000 ml of 0.200 M Fe(NO₃)₃ is combined with 4.000 ml of 0.00200M NaSCN in a 50.00 ml volumetric flask . The flask is then filled to the mark with water. What is the diluted concentration for the SCN^- ion, after the solution is made but before any reaction occurs.

Answer 1

molarity of SCN^- = moles SCN^- ÷ Liters

Find the molarity, so find moles and total volume in liters:

moles of SCN^-

mL to moles, using molarity:

(4.000mL NaSCN / 1 ) × (0.00200 mol NaSCN / 1000 mL NaSCN)=8.0×10^-6 mol NaSCN

mol NaSCN to mol SCN^-

NaSCN has just 1 SCN^-, so:

1 mol NaSCN=1 mol SCN^-, and

8.0×10^-6 mol NaSCN=8.0×10^-6 mol SCN^-

Now find the total volume in Liters:

It's in a 50.0 mL volumetric flask, and

1000mL=1L, so

(^{50.0 mL}/₁)×(^1L/_1000mL)=0.05000 L

Molarity = (8.0×10^-6 mol SCN^-) ÷ 0.05000 L

Molarity = 0.000160 M, or 1.60×10^-4 M