To 110.0 mL of a solution that contains 0.120M Cr(NO3)2 and 0.430M HNO3 is added

Question

To 110.0 mL of a solution that contains 0.120M Cr(NO3)2 and 0.430M HNO3 is added 25.0 mL of 0.240M K2Cr2O7.  The dichromate and Chromium (II) ions react to give Chromium (III) ions.

a)  Write a balance net ionic equation for the reaction.

b)  Calculate concentrations of all ions in the solution after the reaction.  Check to make sure the solution is electrically neutral.

Note:  This has to do with balancing equations for Redox Reactions...balancing H with H+ and O with H20 etc.

Please show alll work:-)

Explanation / Answer

a)Cr2+ ----> Cr3+ + e-

(Cr2O7)2- + 14H+ + 6e- -----> 2Cr3+ + 7H2O

6 Cr2+ +(Cr2O7)2- +14H+ -------> 8 Cr 3+ + 7H2O is the net ionic equation

b)

Moles of Cr 2+ initially = 0.12*0.11 = 0.0132

Moles of HNO3 (H+) initially = 0.43*0.11 = 0.0473

Moles of K2Cr2O7 initially = 0.025*0.24 = 0.006

moles of Cr2O3 = (8/6) *moles of Cr 2+ = (8/6)*0.0132 = 0.0176 moles

Concentration of Cr2O3 = moles/volume = 0.0176/(0.11+0.025) = 0.13 M

moles of H20 formed = (7/6)*moles of Cr 2+ = (7/6)*0.0132 = 0.0154 moles

moles of Cr(NO3)2 remaining =0

Concentration of Cr(NO3)2 =0 M

moles of HNO3 remaining = 0.0473-(0.0132*14/6) = 0.0165 moles

Concentration of HNO3 = 0.0165/(0.11+0.025) = 0.122 M

moles of K2Cr2O7 remaining = 0.006-(0.0132/6) = 0.0038 moles.

Concentration of K2Cr2O7 = 0.0038/(0.11+0.025) = 0.028 M

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.