How many milliliters of 0.325 N HCl are nescessary to reduce 37.6 milliliters of a 0.526 N solution of K2Cr2O7?

Answer 1

First:

Comparing the electropotential values of both redox couples (being equal, see below 1. and 2.)

AND considering that Cl2 gas might escape the mixture giving toxic fumes,

it does NOT look a good titration method.

1. 2Cl- --> Cl2,g + 2e- (delta)Eo= -1.36V
2. Cr2O72- + 14H+ + 6e- --> 2Cr3+ + 7H2O (delta)Eo= +1.36V

Second:

There is no exact answer possible because the titration end point can never be established, you'll never know when the last Cl2 is escaped, and there is no sharp (sudden) change in colour, and the reaction at the endpoint is very slow.

Third: The answer is 60.9 mL

The equivalent point (theorethical end point) can be calculated as follows:

mEq reductor (HCl) = mEq oxidator (K2Cr2O7)0.325 (mEq/mL HCl) * VHCl (mL) = 0.526 (mEq/mL K2Cr2O7) * 37.6 (mL K2Cr2O7)

So VHCl (mL) = 0.526 (mEq/mL K2Cr2O7) * 37.6 (mL K2Cr2O7) / 0.325 (mEq/mL HCl) = 60.854 = 60.9 mL

Answer 2

To calculate the volume needed, we can use the equation M1V1 = M2V2 where M1 is the concentration of HCl, V1 is the volume of HCl, M2 is the concentration of K2Cr2O7, and V2 is the volume of K2Cr2O7. Plugging in the values, we get (0.325)(V1) = (0.526)(37.6). Solving for V1, the volume of 0.325 N HCl needed is approximately 64.52 mL.