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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: jj089 on March 24, 2014, 10:28:23 AM

Title: Calculate the Concentrations of Each Ionic Species
Post by: jj089 on March 24, 2014, 10:28:23 AM

A 35.0 mL aliquot of 0.120 M silver (I) nitrate solution is mixed with 27.0 mL of
0.0750 M potassium chromate solution. Using the solubility table provided, determine
if a precipitate will form and the concentrations of each ionic species at equilibrium.

I determined that a precipitate will form:

Q = [Ag+]^2[CrO42-] = (0.06774)^2(0.03266) = 1.50*10^-4 > Ksp = 1.2*10^-12

How do I calculate the concentrations of each ionic species? Ag+, CrO42-, K+, and NO3-

Title: Re: Calculate the Concentrations of Each Ionic Species
Post by: Borek on March 24, 2014, 11:02:27 AM

What is in excess?

If you assume reaction went to completion - what would be the concentration of the excess reagent?

Title: Re: Calculate the Concentrations of Each Ionic Species
Post by: jj089 on March 24, 2014, 01:22:49 PM

I already know how to find [K+] and [NO3-].

I find that Ag+ is in excess:

0.0677 M Ag+ * 62.0 mL / 1000 mL = 0.00420 mol Ag+

0.0327 M CrO42- * 62.0 mL / 1000 mL = 0.00203 mol CrO42-

2 moles of Ag+ reacts with 1 mol CrO42-

Therefore,

[CrO42-]eq = 0.00 M

[Ag+]eq = [(0.00420/2) - 0.00203][1000]/62.0 = 1.21*10^-3 M. 

Did I do it right?

Title: Re: Calculate the Concentrations of Each Ionic Species
Post by: Borek on March 24, 2014, 01:27:34 PM

Close, but you are off by the factor of 2.