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+ ]?

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

where is the following

explanation:

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