Using the appropriate Ksp values, find the concentration of Cu2+ ions in the sol

Question

Using the appropriate Ksp values, find the concentration of Cu2+ ions in the solution at equilibrium after 700 mL of 0.50 M aqueous Cu(NO3)2 solution has been mixed with 450 mL of 0.25 M aqueous KOH solution. (Enter in M.) (Ksp for Cu(OH)2 is 2.6x10-19).

Now find the concentration of OH ions in this solution at equilibrium. (Enter in M.)

If 250 mL of some Pb(NO3)2 solution is mixed with 450 mL of 7.70 x 102 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 105 M at this temperature.) Enter the concentration in M.

The most important commercial process for generating hydrogen gas is the water-gas shift reaction:

CH4(g) + H2O(g) --> CO(g) + 3 H2(g)

Use the tabulated thermodynamic data to find Go, in kJ. (enter your answer to 4 significant figures)

Now calculate Go1250 for this process when it occurs at 1250 K, in kJ. (enter your answer to 4 significant figures)

Explanation / Answer

Q: Using the appropriate Ksp values, find the concentration of Cu2+ ions in the solution at equilibrium after 700 mL of 0.50 M aqueous Cu(NO3)2 solution has been mixed with 450 mL of 0.25 M aqueous KOH solution. (Enter in M.) (Ksp for Cu(OH)2 is 2.6x10-19). Now find the concentration of OH- ions in this solution at equilibrium. (Enter in M.)

Answer: The reaction involved here is:

    Cu(NO3)2 + 2 KOH -------> Cu(OH)2 + 2 KNO3

The first step is to identify the initial moles:

Moles of Cu(NO3)2 = 0.700 L * 0.50 M (mol/L) = 0.350 moles of Cu(NO3)2 ions
Moles of KOH ions = 0.450 L * 0.25 M (mol/L) = 0.1125 moles of KOH ions

Per the above reaction, for every 0.1125 moles of KOH ions, only 1/2*(0.1125 moles) = 0.05625 of Cu(NO3)2 are required. So KOH is the limiting reactant here. And 0.1125 moles of KOH ions produce 0.05625 moles of Cu(NO3)2

We can setup the ICE table as follows:

Cu(NO3)2 + 2 KOH -------> Cu(OH)2 + 2 KNO3

Initial 0.350 moles      0.1125 moles 0 0

Change -0.05625 moles - 0.1125 moles +0.05625 moles + 0.1125 moles

Equilibrium 0.29375 moles 0 moles   +0.05625 moles + 0.1125 moles

1 mole of Cu2+ ion is produced for every 1 mole of Cu(NO3)2.As 0.29375 moles of Cu(NO3)2 areleft over in the solution, 0.29375 moles of Cu2+ ions would remain unreacted. And the total volume is 0.700 L + 0.450 L = 1.150 L

[Cu2+] = moles of Cu2+/Vol in L = 0.29375/1.150 = 0.255 M

Moles of [OH-] at equilibrium = moles of KOH at equilibrium = 0.05625

[OH-] = moles of OH-/Vol in L = 0.05625/1.150 = 0.0489 M

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