[Aisha_18]

have to find the molarity of an Unknown acid, but I was wondering how.
I used this equation
Cr2O7+14H+ + 6S2O3 = 2Cr + 3S4O6 + 7h2O
We had to find the molarity of sodium thiosulphate which was 0.1M and K2Cr2O7 which was 0.02M for part A

Part B
This was an idometric titration the volumes used were 25 cm 3 of potassium iodate, 10cm3 of potassium iodide, starch solution as an idicator and then 23.09 cm3 of sodium thiosulphate.
Since the acid was unknown how would I know what ratio it is compared to the other known moles? To find the molarity

[Perito]

Part A:

Dichromate reacts with iodide to yield iodine.

Cr2O7(-2) + 6 I(-) + 14H+ → 2 Cr2(+3) + 7 H2O + 3 I2

Acid and iodide are usually added in excess. Thus, one dichromate ion produces three iodine molecules.

Note that when you balance an equation, you must also balance the charge (or show the counterions). In this case, we have "-2 -6 +14 = +6" on the left and +6 on the right, so the charges balance.

The iodine can be titrated with thiosulfate to yield iodide. When all of the iodine has been consumed, the indicator (starch) turns from deep blue to colorless. It's a very easy-to-see endpoint.

I2 + 2 S2O3(2−) → S4O6(2−) + 2 I−

One mole of iodine reacts with two moles of thiosulfate.

Part B. You have a potassium iodate solution (KIO3). Iodate will react with iodide to form Iodine:

IO3- + 5I- +6 H+ = 3 I2 + 3 H2O

If the acid is limiting, as it appears to be in this case, you can titrate the I2 with thiosulfate and back-calculate the amount of acid. You can't tell the molarity of the acid because you don't know if it's a mono-protic acid, a di-protic acid (two hydrogen ions), etc. but you can tell the normality of the acid since that is not dependent on the type of acid.