Problem 127 - Chemistry

Question

Place the species in each of the following groups in order of increasing acid strength. Explain the order you chose for each group. a. \(\mathrm{HIO}_{3}, \mathrm{HBrO}_{3}\) c. HOCl, HOI b. \(\mathrm{HNO}_{2}, \mathrm{HNO}_{3}\) d. \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{3} \mathrm{PO}_{3}\)

Step-by-Step Solution

Answer

The order of increasing acid strength for each group is: a. \(\mathrm{HIO}_{3} < \mathrm{HBrO}_{3}\), because \(\mathrm{HBrO}_{3}\) forms a more stable anion (\(\mathrm{BrO}_{3}^-\)) than \(\mathrm{HIO}_{3}\). c. HOI < HOCl, because HOCl forms a more stable anion (\(\mathrm{ClO}^-\)) than HOI. b. \(\mathrm{HNO}_{2} < \mathrm{HNO}_{3}\), because \(\mathrm{HNO}_{3}\) forms a more stable anion (\(\mathrm{NO}_{3}^-\)) than \(\mathrm{HNO}_{2}\). d. \(\mathrm{H}_{3} \mathrm{PO}_{3} < \mathrm{H}_{3} \mathrm{PO}_{4}\), because \(\mathrm{H}_{3} \mathrm{PO}_{4}\) forms a more stable anion (\(\mathrm{H}_{2} \mathrm{PO}_{4}^-\)) than \(\mathrm{H}_{3} \mathrm{PO}_{3}\).

Step 1: Identify central atoms and electronegativity trends

For both \(\mathrm{HIO}_{3}\) and \(\mathrm{HBrO}_{3}\), the central atom is a halogen (iodine and bromine, respectively). When moving down the periodic table, electronegativity decreases. Therefore, iodine is less electronegative than bromine.

Step 2: Analyze the stability of the anions formed after donating a proton

After donating a proton (H+), \(\mathrm{HIO}_{3}\) forms \(\mathrm{IO}_{3}^-\) and \(\mathrm{HBrO}_{3}\) forms \(\mathrm{BrO}_{3}^-\). Since iodine is less electronegative than bromine, the negative charge on the iodate ion (\(\mathrm{IO}_{3}^-\)) will be less stable. Less stable ions tend to form weaker acids.

Step 3: Order the species in increasing acid strength

Since \(\mathrm{HIO}_{3}\) forms a less stable ion than \(\mathrm{HBrO}_{3}\), its acid strength is lower. Therefore, the order of increasing acid strength is \(\mathrm{HIO}_{3} < \mathrm{HBrO}_{3}\). c. HOCl, HOI

Step 1: Identify central atoms and electronegativity trends

For both HOCl and HOI, the central atom is a halogen (chlorine and iodine, respectively). Chlorine is more electronegative than iodine.

Step 2: Analyze the stability of the anions formed after donating a proton

After donating a proton, HOCl forms \(\mathrm{ClO}^-\) and HOI forms \(\mathrm{IO}^-\). Since chlorine is more electronegative than iodine, the negative charge on the hypochlorite ion (\(\mathrm{ClO}^-\)) will be more stable.

Step 3: Order the species in increasing acid strength

Since HOCl forms a more stable ion than HOI, its acid strength is higher. Thus, the order of increasing acid strength is HOI < HOCl. b. \(\mathrm{HNO}_{2}, \mathrm{HNO}_{3}\)

Step 1: Compare the structures of the species

Both species have nitrogen and oxygen. The difference lies in the number of oxygens directly bonded to nitrogen: \(\mathrm{HNO}_{2}\) has one, whereas \(\mathrm{HNO}_{3}\) has two.

Step 2: Analyze the stability of the anions formed after donating a proton

After donating a proton, \(\mathrm{HNO}_{2}\) forms \(\mathrm{NO}_{2}^-\) and \(\mathrm{HNO}_{3}\) forms \(\mathrm{NO}_{3}^-\). More oxygens in \(\mathrm{NO}_{3}^-\) allow for better delocalization of the negative charge, making it more stable.

Step 3: Order the species in increasing acid strength

Since \(\mathrm{HNO}_{3}\) forms a more stable ion than \(\mathrm{HNO}_{2}\), its acid strength is higher. Thus, the order of increasing acid strength is \(\mathrm{HNO}_{2} < \mathrm{HNO}_{3}\). d. \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{3} \mathrm{PO}_{3}\)

Step 1: Compare the structures of the species

Both species have phosphorus and oxygen. \(\mathrm{H}_{3} \mathrm{PO}_{4}\) has four oxygens bonded to phosphorus, whereas \(\mathrm{H}_{3} \mathrm{PO}_{3}\) has three oxygens bonded to phosphorus.

Step 2: Analyze the stability of the anions formed after donating a proton

After donating a proton, \(\mathrm{H}_{3} \mathrm{PO}_{4}\) forms \(\mathrm{H}_{2} \mathrm{PO}_{4}^-\) and \(\mathrm{H}_{3} \mathrm{PO}_{3}\) forms \(\mathrm{H}_{2} \mathrm{PO}_{3}^-\). The extra oxygen in \(\mathrm{H}_{2} \mathrm{PO}_{4}^-\) allows for better charge delocalization, making it more stable.

Step 3: Order the species in increasing acid strength

Since \(\mathrm{H}_{3} \mathrm{PO}_{4}\) forms a more stable ion than \(\mathrm{H}_{3} \mathrm{PO}_{3}\), its acid strength is higher. Thus, the order of increasing acid strength is \(\mathrm{H}_{3} \mathrm{PO}_{3} < \mathrm{H}_{3} \mathrm{PO}_{4}\).