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amit25
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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 :(