Calculating Concentrations of CrCl(OH2)5+2 and Cr(OH2)6+3

[amit25]

Homework Statement

10 mL solution of 0.5 M CrCl(OH2)52+ is allowed to aquate to Cr(OH2)63+. To determine an approximate rate of reaction, the amounts of CrCl(OH2)52+ and Cr(OH2)63+ present after a certain period of time are evaluated. This is done by pouring the solution on a cation exchange resin in the H+ form and then titrating the displaced H+ with base. If 80 mL of 0.15 M NaOH is required to neutralize the liberated H +, what was the concentration of CrCl(OH2)52+ and Cr(OH2)63+ in the solution?

Homework Equations

The Attempt at a Solution

CrCl)OH2)5 +2

V=10ml c=0.5M
n=(0.01L)(0.5M)
n=0.005mol

NaOH

V=80mL c=0.15M
n=(0.15M)(80/1000)
n=0.012mol

Total volume=90mL
c1v1=c2v2
(0.5M)(10mL)=c2(90ml)
c2=0.056M

im not really sure if I am doing this right, any help please :(

 

[Borek]

I don't think you are doing it right, even if you have not wrote what you are doing and what c1 and c2 are.

How many moles of H⁺ are liberated per each mole of CrCl(H₂O)₅²⁺?

How many moles of H⁺ are liberated per each mole of Cr(H₂O)₆³⁺?

If a mixture contains n₁ moles of CrCl(H₂O)₅²⁺ and n₂ moles of Cr(H₂O)₆³⁺, how many moles of H⁺ will be liberated? How many moles of NaOH will be needed to titrate the elute?

And finally, what does n₁+n₂ equal to?

This gives you two equations in two unknowns and finding the concentrations of both complexes shouldn't be difficult.