Assuming each solution to be 0.10 M , rank the following aqueous solutions in order of decreasing pH. Rank the solutions from the highest to lowest pH. To rank items as equivalent, overlap themN2H2Ba(OH)2HOCLNAOHHCL

Ba(OH)₂ > NaOH >N₂H₂  > HOCl > HCl

Explanation:

To get a better understanding of this, we need to know what's the meaning of the pH values.

The pH value of any compound basically gives us information of how acidic or basic is a compound. If this value is higher (>7), means major basicity; if this number is lower (<7) means major acidity in the compound.

According to this, a strong acid would be the one with the lowest pH, and a strong base would be the one with the highest pH.

Among the 5 compounds we have here, only 2 of them are acid, and the rest is basic, so we can assume that both acid would have the lower pH, it's up to us to see which of them have the lowest pH.

To do that, we can use the following expression:

pH = -log[H⁺]

If all solutions have a concentration of 0.10, then the pH would be:

HCl:

pH = -log(0.1) = 1

HOCL:

This one is a weak acid, so it's not completely dissociated in solution and the [H⁺] would be <0.1 M. As its a weak acid, we can assume that it's pH would be near 4, 5 or 6. We cannot know with all certain, because we need the Ka of this acid, and see the dissociation of the acid in solution, but for this case, it's fine and HCl is one of the strongest known acid.

In this case the order is: HOCl > HCl

For the bases, we have 3 bases, 2 of them are strong bases, and 1 of them is a weak base so it doesn't dissociate completely (N₂H₂), so this would have the lowest pH among the bases. In the case of bases, we calculate pOH, (Same formula as before) and then, we substract 14 from the value of pOH. As for the other two:

Ba(OH)₂:

Concentration would be double because we have 2 OH, so:

[OH⁻] = 2 * 0.1 = 0.2 M

pOH = -log(0.2) = 0.69

pH = 14 - 0.69 = 13.31

NaOH:

pOH = -log(0.1) = 1

pH = 14 - 1 = 13

Therefore, Ba(OH)₂ is stronger than NaOH and N₂H₂. So, our order would be:

Hope this helps