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

  • Thread starter amit25
  • Start date
In summary, a solution of 10 mL of 0.5 M CrCl(OH2)52+ was allowed to aquate to Cr(OH2)63+, and the amounts of each complex present after a certain period of time were evaluated using a cation exchange resin and titration with base. The concentration of CrCl(OH2)52+ and Cr(OH2)63+ in the solution was found to be approximately 0.056 M. Two equations in two unknowns were used to find the concentrations, and the total number of moles of both complexes was determined.
  • #1
amit25
30
0

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 :(
 
Physics news on Phys.org
  • #2
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.
 

What is the formula for calculating concentrations of CrCl(OH2)5+2 and Cr(OH2)6+3?

The formula for calculating concentrations of CrCl(OH2)5+2 and Cr(OH2)6+3 is C = n/V, where C is the concentration in mol/L, n is the number of moles of the compound, and V is the volume in liters.

How do you determine the number of moles of CrCl(OH2)5+2 and Cr(OH2)6+3 in a solution?

To determine the number of moles, you need to know the molar mass of the compounds and the mass or volume of the solution. You can use the formula n = m/M, where n is the number of moles, m is the mass in grams, and M is the molar mass in g/mol.

Can you explain the difference between CrCl(OH2)5+2 and Cr(OH2)6+3?

The difference between CrCl(OH2)5+2 and Cr(OH2)6+3 is the charge of the central chromium ion. In CrCl(OH2)5+2, the chromium ion has a +2 charge, while in Cr(OH2)6+3, the chromium ion has a +3 charge. This difference in charge affects the properties and reactivity of the compounds.

What is the role of water molecules in these compounds?

The water molecules in these compounds act as ligands, meaning they bond to the central chromium ion. This affects the overall charge and structure of the compounds. In aqueous solutions, the water molecules also play a role in determining the concentration of the compounds.

How does temperature affect the concentrations of CrCl(OH2)5+2 and Cr(OH2)6+3?

Temperature can affect the concentrations of these compounds by changing the volume of the solution. As temperature increases, the volume of the solution also increases, leading to a decrease in concentration. Additionally, changes in temperature can also affect the solubility of these compounds, which can impact the concentration.

Similar threads

Replies
2
Views
5K
  • Biology and Chemistry Homework Help
Replies
4
Views
10K
Back
Top