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If 200 mL of some Pb(NO3)2 solution is mixed with 500 mL of 3.50 x 10−2 M NaCl solution, what is the maximum concentration of the Pb(NO3)2 solution added if no solid PbCl2 forms? (Assume Ksp = 2.00 x 10−5 M at this temperature.) Enter the concentration in M.

Question

If 200 mL of some Pb(NO3)2 solution is mixed with 500 mL of 3.50 x 10−2 M NaCl solution, what is the maximum concentration of the Pb(NO3)2 solution added if no solid PbCl2 forms? (Assume Ksp = 2.00 x 10−5 M at this temperature.) Enter the concentration in M.

check_circleAnswer
Step 1

Total volume of solution is 200 + 500 = 700 mL = 0.7L

Moles of chloride ions (Cl-) = 0.500*3.50*10-2 = 0.0175

The concentration of chloride ions (Cl-) can be calculated as

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Step 2

We can write the equilibrium expression for t...

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Chemistry

Equilibrium Concepts

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