Question

Given the following reagents: 15.50 ml of 0.335 MLead(IV) Nitrate   and   25.00ml of 0.400MSodium Sulfide, If they were mixed:

1. What is the resulting concentration of the Lead ion?
2. What is the resulting concentration of the Sodium ion?
3. What is the resulting concentration of the Nitrate ion?

Answer 1

Molarity=Number of moles of solute/Volume of solution (L)

Number of moles of solute=Molarity x Volume (L)

Volume of Lead(IV) nitrate solution=15.50 mL=15.50 mL/1000 mL/L=0.01550 L (1 L=1000 mL)

Molarity of Lead (IV) nitrate solution=0.335 M

Number of moles of Lead (IV) nitrate=0.335 M x 0.01550 L=0.0051925 mol0.005 mol

Lead (IV) nitrate dissociates as shown

So 1 mol Lead (IV) nitrate gives 1 mol Lead (IV) ions and 4 mol nitrate ions in solution

0.005 mol Lead (IV) nitrate gives 0.005 mol Lead (IV) ions and 4x0.005=0.020 mol nitrate ions in solution

Volume of Sodium sulfide solution=25.00 mL=25.00 mL/1000 mL/L=0.025 L

Molarity of Sodium sulfide solution=0.400 M

Number of moles of sodium sulfide=0.004 M x 0.025 L=0.010 mol

Sodium sulfide dissociates as shown

1 mol sodium sulfide gives 2 mol sodium ions in solution

0.010 mol sodium sulfide gives 2x0.010=0.020 mol sodium ions in solution

Total final volume after mixing up the two solutions=0.01550 L+0.025 L=0.0405 L

A) Resulting concentration of Lead (IV) ions=number of moles of Lead (IV) ions/Total volume of solution (L)=0.005 mol/0.0405 L=0.123 M

B) Resulting concentration of sodium ions=number of moles of sodium ions/Total volume of the solution (L)=0.020 mol/0.405 L=0.494 M

C)Resulting concentration of nitrate ions=number of moles of nitrate ions/Total volume of solution (L)=0.020 mol/0.0405 L=0.494 M