[amit25]

10mL 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?

[AWK]

You should show your work.
Start from balanced reaction of both form with cation exchange resin (separately).

[amit25]

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 im doing this right, any help please

[amit25]

Cr(H2O)₄Cl2+ + 2H₂O  Cr(H₂O)₅ Cl2+ + Cl- + H₂O  Cr(H2O)₆ ³⁺ + 2 Cl-
and this is the balanced equation?

[AWK]

You should write down reactions that liberating H⁺ and do their stoichiometry

[amit25]

H₂O + CrCl(OH2)5 +2  Cr(OH2)6 +3    + Cl-
 
so the ratio is 1:1 for CrCl(OH2)5 +2  and  Cr(OH2)6 +3

the moles for each is n=(0.5M)(0.01L)=0.005mol

??