Question

a. write the balanced equation, including phase symbols, for the reaction between hi and manganese (iii) hydroxide.
3hi(aq) + mn(oh)₃(s) → mni₃(aq) + 3h₂o(l)
b. calculate the volume of 0.7872 m mn(oh)₃ needed to completely neutralize 11.73 ml of 0.5196 m hi.

Explanation:

Step1: Recall Molarity Formula

The formula for molarity is \( M = \frac{n}{V} \), where \( M \) is molarity, \( n \) is moles, and \( V \) is volume in liters. For neutralization, moles of acid = moles of base (from balanced equation). From part A, the ratio of \( HI \) to \( Mn(OH)_3 \) is 3:1.

Step2: Calculate Moles of HI

Moles of \( HI \), \( n_{HI} = M_{HI} \times V_{HI} \). \( M_{HI} = 0.5196 \, M \), \( V_{HI} = 11.73 \, mL = 0.01173 \, L \). So \( n_{HI} = 0.5196 \times 0.01173 \).

Step3: Find Moles of \( Mn(OH)_3 \)

From ratio 3:1, \( n_{Mn(OH)_3} = \frac{n_{HI}}{3} \).

Step4: Calculate Volume of \( Mn(OH)_3 \)

Using \( V = \frac{n}{M} \), \( M_{Mn(OH)_3} = 0.7872 \, M \). So \( V_{Mn(OH)_3} = \frac{n_{Mn(OH)_3}}{0.7872} \).

First, calculate \( n_{HI} = 0.5196 \times 0.01173 = 0.006095908 \, mol \).

Then, \( n_{Mn(OH)_3} = \frac{0.006095908}{3} \approx 0.002031969 \, mol \).

Finally, \( V = \frac{0.002031969}{0.7872} \approx 0.002581 \, L = 2.581 \, mL \).

Answer:

\boxed{2.581 \, mL} (or 2.58 mL, depending on significant figures)