A 1.458 g of Mg reacts with 80.0 ml of a HCl solution whose pH is -0.477.The change in pH when all Mg has reacted.(Assume constant volume. `Mg=24.3g//"mol".`)(log 3=0.47, log2= 0.301)

Options

A. -0.176
B. `+0.477`
C. -0.2345
D. 0.3

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Write the Balanced Chemical Equation
The reaction between magnesium (Mg) and hydrochloric acid (HCl) can be represented as:
\[ \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \]

### Step 2: Calculate the Moles of Magnesium
To find the number of moles of magnesium, we use the formula:
\[ \text{Moles of Mg} = \frac{\text{mass of Mg}}{\text{molar mass of Mg}} \]
Given:
- Mass of Mg = 1.458 g
- Molar mass of Mg = 24.3 g/mol

Calculating the moles:
\[ \text{Moles of Mg} = \frac{1.458 \text{ g}}{24.3 \text{ g/mol}} \approx 0.060 \text{ mol} \]

### Step 3: Convert Moles of Mg to Millimoles
To convert moles to millimoles:
\[ 0.060 \text{ mol} = 60 \text{ mmoles} \]

### Step 4: Calculate the Molarity of HCl
Given that the volume of HCl solution is 80.0 mL, we need to find the molarity of HCl. The pH of the solution is given as -0.477. We can calculate the concentration of H⁺ ions using:
\[ \text{pH} = -\log[\text{H}^+] \]
Thus,
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{0.477} \approx 3 \text{ M} \]

### Step 5: Calculate Total Moles of HCl
Using the molarity calculated:
\[ \text{Total moles of HCl} = \text{Molarity} \times \text{Volume (L)} \]
Convert 80 mL to liters:
\[ 80 \text{ mL} = 0.080 \text{ L} \]
Now calculate:
\[ \text{Total moles of HCl} = 3 \text{ mol/L} \times 0.080 \text{ L} = 0.240 \text{ mol} = 240 \text{ mmoles} \]

### Step 6: Calculate Remaining HCl After Reaction
From the balanced equation, 1 mole of Mg reacts with 2 moles of HCl. Therefore, 60 mmoles of Mg will react with:
\[ 60 \text{ mmoles Mg} \times 2 = 120 \text{ mmoles HCl} \]
Remaining HCl after the reaction:
\[ \text{Remaining HCl} = 240 \text{ mmoles} - 120 \text{ mmoles} = 120 \text{ mmoles} \]

### Step 7: Calculate New Molarity of HCl
The new molarity of HCl after the reaction is:
\[ \text{New Molarity} = \frac{\text{Remaining moles of HCl}}{\text{Volume (L)}} = \frac{120 \text{ mmoles}}{80 \text{ mL}} = \frac{120}{0.080} = 1.5 \text{ M} \]

### Step 8: Calculate the New pH
Using the new molarity to find the new pH:
\[ \text{pH} = -\log(1.5) \]
Using the logarithm values provided:
\[ \text{pH} \approx -\log(1.5) \approx -0.176 \]

### Step 9: Calculate the Change in pH
The initial pH was -0.477, and the new pH is approximately -0.176. The change in pH is:
\[ \Delta \text{pH} = \text{New pH} - \text{Initial pH} = -0.176 - (-0.477) \]
Calculating this gives:
\[ \Delta \text{pH} = 0.301 \]

### Final Answer
The change in pH when all Mg has reacted is approximately **0.301**.

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