[Lily25]

10 mL solution of 0.5 M CrCl(OH2)5 +2 is allowed to aquate to Cr(OH2)6 +3

To determine an approximate rate of reaction, the amounts of CrCl(OH2)5 +2 and Cr(OH2)6 +3
Present after a certain time are measured by pouring the solution onto 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 were the concentrations of CrCl(OH2)5 2+ and Cr(OH2)6 3+ in the solution

[JudyKayTee]

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[Lily25]

k this is how I did it... help please

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