I just need help with the las question. At first I got 6.5e-5 then I got 1.69e-8

Question

I just need help with the las question. At first I got 6.5e-5 then I got 1.69e-8 then 0. I would really appreciate an explanation and a step by step walk through. Thank you! 5. 18/23 points | Previous Answers Mixing the following solutions resulted in the isolation of 1.61 g of Ag2CrO4 55.00 mL of 0.192 M AgNO3 44.00 mL of 0.157 M K2CrO4 Complete the following reaction table in millimoles 2Ag+Cro42Ag2CrO4 Ksp 1.1e-1:2 initial 10.56 delta -10.56 final 0 What is the theoretical yield of Ag2Cr04 6.908 mmol mmol -5.280 5.280 1.628 5.280 mmo 1.75 What is the percent yield of Ag2CrO4 92 Assume additive volumes to determine the equilibrium concentration of the excess reactant. 0164 Determine the equilibrium concentration of the limiting reactant. Submit Answer Save Progress

Explanation / Answer

Answer

8.19×e-6 M

Explanation

Solubility equillibrium of Ag2CrO4 is

Ag2CrO4(s) <------> 2Ag+(aq) + CrO42-(aq)

Ksp = [Ag+]2[CrO42-] = 1.1×10-12

at equillibrium,

[Ag+]= 2x

[CrO42-]= 0.0164 + x

Therefore,

(2x)2× (0.0164 +x) = 1.1×10-12

Solving for x

x = 4.0944×10-6

Therefore,

[Ag+] =2× 4.0944×10-6 = 8.19×10-6M

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