A 115.0 mL sample of a solution that is 2.6×10^3 M inAgNO3 is mixed with a 230.0

Question

A 115.0 mL sample of a solution that is 2.6×10^3 M inAgNO3 is mixed with a 230.0 mL sample of a solution that is 0.12 Min NaCN.

After the solution reaches equilibrium, what concentration ofAg+(aq) remains?

I double check and this is ALL the information that is given.However, if you need the Kf = 1x10^21. This is what you have tofind yourself. I tried this problem but i am still getting thewrong answer.

Explanation / Answer

milli moles of AgNO3 taken = 115*2.6*10^-3 = 0.299 milli moles of NaCN taken = 23 *0.12 = 2.76 reaction taking place: AgNO3 + NaCN -> NaNO3 +AgCN (white precipitate) 0.299      2.76          0                 0 [intial milli moles] 0        2.461             0.299      0.299       [milli moles at equilibrium] Clearly concentration of Ag+ (aq) is zero, that means it is infinitely small so it is taken as zero To find exact value of Ag+(aq) one has to use the Equilibriumconstant(K) for this reaction, but in this description it isnot given ...? if you can post the K value exact value of Ag+(aq) can becalculated .

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