Chemistry Stoichiometry?
A sample of 1.05 g of lead (II) nitrate is mixed with 122 mL of 0.110 M sodium sulfate solution.
What is the concentration of \rm SO_4^{2-} ion that remain in solution after the reaction is complete?
A sample of 1.05 g of lead (II) nitrate is mixed with 122 mL of 0.110 M sodium sulfate solution.
What is the concentration of \rm SO_4^{2-} ion that remain in solution after the reaction is complete?
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Verified answer
1.05 g / 331.2 g mol-1= 3.17 mmoles lead (II) nitrate.
0.110 mol L-1 x (122 / 1000) L = 13.42 mmoles sodium sulfate.
13.42 - 3.17 = 10.25 mmoles sodium sulfate remaining.
Lead (II) nitrate density = 4.53 g / mL. d = m/v so v = m/d = 1.05 / 4.53 = 0.23 mL which is negligible.
10.25 mmoles in 122 mL = 10.25 / 122 = 0.084 moles / L sulfate ions remaining.