Question

In run 1, you mix 8.0 mL of the 43 g/L MO solution (MO molar mass is 327.33 g/mol), 2.60 mL of the 0.040 M SnCl2 in 2 M HCl solution, 5.43 mL of 2.0M HCl solution, and 3.73 mL of 2.0M NaCl solution. What is the [Sn2+ ]?

Answer 1

The concentration of [Sn⁺²] will be calculated by first calculating the moles of SnCl₂ added as these moles will give us the moles of [Sn⁺²] ion.

Moles of SnCl₂ = molarity X volume = 0.04 X 2.60 = 0.104 milli moles [as volume is in mL]

The moles of [Sn⁺² = 0.104 mmol

the total volume in solution = volume due to MO + volume due to SnCl₂ + volume due to HCl + volume due to NaCl

Total volume = 8+2.60+5.43+3.73= 19.76 mL

Concentration = moles / volume

concentration [Sn⁺²]  = 0.104mmol / 19.76 mL = 0.0053 mol / L