RZeal

Consider a 0.42 M solution of NaCH3COO CH3COO–(aq) + H2O(l) -> CH3COOH(aq) + OH–(aq) Calculate the pH of the solution at equilibrium. Try the weak acid or weak base approximation (as appropriate) and check the validity of the approximation. If the approxi...

Calculate the Ecell for this reaction 3HClO2(aq) +2Cr3+(aq) + 4H2O(l) -> 3HClO(aq) + (Cr2O7)2–(aq) + 8H+(aq) I thought Cr + e -> Cr2O7 is the cathode and 3HClO2 -> 3HClO + e- is the anode. Using Ecell = Cathode - Anode I get -0.31, but the answer I checke...

3HClO2(aq) +2Cr3+(aq) + 4H2O(l) -> 3HClO(aq) + (Cr2O7)2–(aq) + 8H+(aq) At pH 0.00, with [Cr2O72–] = 0.80 M, [HClO2] = 0.15 M, and [HClO] = 0.20 M, the cell voltage is found to be 0.15 V. Calculate the concentration of [Cr3+] in the cell Ecell = 0.31 Not s...

Nevermind I figured it out, With the equation Ka = x^2 / 0.0165 - x x = [H+] I forgot to do the 0.0165-x in my original calculation, but after doing that I got the correct answer.

Ka = Ch3COO- 5.6x10^-10 Ka = Ch3COOH 1.8x10^-5

Sorry about the confusion Ecell = 0.15 and Eocell = 0.31

Thanks I was able to solve it!