What is the pH and pOH of a solution made by adding water to 15 grams of hydroiodic acid until the volume of the solution is 2500 mL? (hint: you must calculate the molarity first)
The Correct Answer and Explanation is:
To solve for the pH and pOH of the solution, we will follow these steps:
Step 1: Find the molarity of the hydroiodic acid (HI) solution
First, we need to calculate the number of moles of HI in 15 grams. The molecular weight of HI is:
- Iodine (I) = 126.9 g/mol
- Hydrogen (H) = 1.008 g/mol
- Therefore, the molar mass of HI = 126.9 + 1.008 = 127.908 g/mol
Now, calculate the moles of HI:Moles of HI=mass of HI (grams)molar mass (g/mol)=15 g127.908 g/mol≈0.1177 mol\text{Moles of HI} = \frac{\text{mass of HI (grams)}}{\text{molar mass (g/mol)}} = \frac{15 \, \text{g}}{127.908 \, \text{g/mol}} \approx 0.1177 \, \text{mol}Moles of HI=molar mass (g/mol)mass of HI (grams)=127.908g/mol15g≈0.1177mol
Next, calculate the molarity of the solution. Molarity is defined as moles of solute divided by the volume of the solution in liters:Molarity=0.1177 mol2.5 L=0.04708 M\text{Molarity} = \frac{0.1177 \, \text{mol}}{2.5 \, \text{L}} = 0.04708 \, \text{M}Molarity=2.5L0.1177mol=0.04708M
Step 2: Calculate the pH of the solution
Since HI is a strong acid, it dissociates completely in water:HI→H++I−\text{HI} \rightarrow \text{H}^+ + \text{I}^-HI→H++I−
The concentration of hydrogen ions [H⁺] is the same as the molarity of the acid, 0.04708 M. To find the pH, we use the formula:pH=−log[H+]\text{pH} = -\log[\text{H}^+]pH=−log[H+]pH=−log(0.04708)≈1.327\text{pH} = -\log(0.04708) \approx 1.327pH=−log(0.04708)≈1.327
Step 3: Calculate the pOH of the solution
The relationship between pH and pOH is:pH+pOH=14\text{pH} + \text{pOH} = 14pH+pOH=14
So:pOH=14−pH=14−1.327=12.673\text{pOH} = 14 – \text{pH} = 14 – 1.327 = 12.673pOH=14−pH=14−1.327=12.673
Final Answer:
- pH ≈ 1.33
- pOH ≈ 12.67
Explanation:
The pH and pOH values are derived from the concentration of hydrogen ions (H⁺) and hydroxide ions (OH⁻), respectively. The solution in question is strongly acidic due to the dissociation of hydroiodic acid, a strong acid. By calculating the molarity of HI and recognizing that it dissociates completely, we can directly relate the molarity to the concentration of hydrogen ions. Then, applying the formula for pH and using the relationship between pH and pOH, we obtain the final values of 1.33 for pH and 12.67 for pOH.
