Question

calculate the volume of 0.150 m hydrochloric acid required to completely neutralize 25.0 ml of 0.250 m iron (iii) hydroxide.

Answer 1

0.125 L HCl solution, or 125 mL HCl solution (Depending on the units requested)

Step-by-step explanation:

Major steps:

1. Determine the chemical formulas for each compound

2. Write the unbalanced chemical equation, and balance it

3. Use dimensional analysis to determine the amount of acid needed.

Step 1. Determine the chemical formulas for each compound

hydrochloric acid is
`HCl`. This is from memorization of nomenclature, or consulting a resource.

Iron (iii) hydroxide is
`Fe(OH)_3` . This is from memorization of nomenclature, knowing that the charge on "hydroxide" is a negative 1, and that 3 hydroxide ions will be needed to balance the charge with a Iron (iii), or consulting a resource.

Step 2. Write the unbalanced chemical equation, and balance it

For "neutralization reactions", an "Acid" and a "Base" will combine to form Water and a "salt".

Unbalanced chemical equation:

`HCl + Fe(OH)_(3) \rightarrow H_(2)O+ FeCl_(3)`

Balance the equation by increase the number of "Chlorines" on the left, and the number of "hydroxides" (trapped in the 'water') on the right.

Balanced chemical equation:

`3HCl + Fe(OH)_(3) \rightarrow 3H_(2)O+ FeCl_(3)`

Step 3. Use dimensional analysis to determine the amount of acid needed.

Knowing we have 25.0mL of Iron (iii) hydroxide solution (in milliliters), we first convert to Liters (since concentrations for "molarity" are measured in moles per Liter).

Then convert to convert to moles of Iron(iii) hydroxide using the solution's concentration.

Convert to moles of hydrochloric acid using the mole ratio from the balanced chemical equation.

Lastly convert to volume of the hydrochloric acid solution using that solution's concentration:

`\frac{25.0 \text{ mL } Fe(OH)_3 \text{ solution}}{1} * \frac{1 \text{ L }}{1000 \text{ mL }} * \frac{0.25 \text{ mol } Fe(OH)_3 }{1 \text{ L } Fe(OH)_3 \text{ solution}} * \frac{3 \text{ mol } HCl }{1 \text{ mol } Fe(OH)_3 } * \frac{1 \text{ L } HCl \text{ solution} }{0.150 \text{ mol } HCl }=`

`=0.125 \text{ L } HCl \text{ solution}`

If the requested answer should be measured in milliliters, one last conversion will yield the answer:

`\frac{0.125 \text{ L } HCl \text{ solution}}{1} * \frac{1000 \text{ mL }}{1 \text{ L }} = 125 \text{ mL } HCl \text{ solution}`

Observe that the original measurements use 3 significant figures, so each answer should use 3 significant figures (both answers do).