Question

The pH of a solution prepared by mixing 40.0 mL of 0.125 M Mg(OH)2 and 150.0 mL of 0.125 M HCl is __________.

A) 6.29
B) 4.11
C) 1.14
D) 5.78
E) 1.34

Answer 1

The pH of the solution prepared by mixing 40.0 mL of 0.125 M Mg(OH)2 and 150.0 mL of 0.125 M HCl is approximately 1.34.

Here's a summary of the steps taken:

1. Calculate the moles of hydroxide ions (OH⁻) from Mg(OH)₂. Since each molecule provides two OH⁻ ions, the total moles of OH⁻ is twice the product of the concentration and volume of the Mg(OH)₂ solution.

2. Calculate the moles of hydrogen ions (H⁺) from HCl, which is equal to the product of the concentration and volume of the HCl solution.

3. Determine the limiting reactant by comparing the moles of H⁺ and OH⁻.

4. Calculate the remaining moles of the excess ion (H⁺ or OH⁻) after the reaction.

5. Find the concentration of the excess ion in the combined volume of the solutions.

6. If H⁺ is in excess, calculate pH directly; if OH⁻ is in excess, calculate pOH and then convert to pH using the relation pH + pOH = 14.

7. Round the calculated pH to two decimal places to match the answer choices.

Therefore, the correct answer is:

E) 1.34

Answer 2

The pH of the solution is approximately 1.34. Option E.

To find the pH of the solution after mixing Mg(OH)₂ and HCl, we need to determine the concentration of the remaining hydroxide ions or hydronium ions in the solution.

First, find the moles of Mg(OH)₂ and HCl:

Moles of Mg(OH)₂ = molarity x volume

= 0.125 x 0.004

= 0.005 mol

Moles of HCl = 0.125 x 0.150

= 0.01875 mol

Since the reaction between Mg(OH)₂ and HCl is a 1:2 ratio, it will completely react with the limiting reagent, which is Mg(OH)₂.

`\[ Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O \]`

This means that all of the moles of Mg(OH)₂ will react, leaving some excess HCl unreacted.

Now, find the moles of excess HCl:

`\[ \text{moles of excess HCl} = \text{moles of HCl initially} - (\text{moles of Mg(OH)}₂ * 2) \]`

= 0.01875 - (0.005 x 2)

= 0.00875 mol

Now, find the concentration of
`\(H_3O^+\)` ions produced by the excess HCl:

`\[ \text{Concentration of } H_3O^+ = \frac{\text{moles of excess HCl}}{\text{total volume of solution in liters}} \]`

=
`\frac{0.00875 \, \text{mol}}{0.040 \, \text{L} + 0.150 \, \text{L} } \]`

`\[ \text{Concentration of } H_3O^+ \approx \frac{0.00875 \, \text{mol}}{0.190 \, \text{L}} \]`

= 0.04605 M

Now, find the pH:

`\[ \text{pH} = -\log(0.04605) \]`

`\[ \text{pH} \approx 1.34 \]`

So, the pH of the solution is approximately 1.34.