The question requires us to calculate the pOH value for a solution containing 0.01 moles of HCl in 2.5 L of solution.
To solve this problem, we'll need to go through the following steps:
I) calculate the molar concentration of the HCl solution to obtaine the concentration of H+ ions;
II) using the concentration of H+ ions, calculate the pH of the solution;
III) with the value obtained for pH, calculate the value of pOH for this solution.
Next, we'll go through the steps to solve the question:
I) Since the question provided the number of moles of HCl and the volume of the solution, we can calculate the molar concentration using the following equation:
![\text{molar concentration = }\frac{\text{number of moles (mol)}}{\text{volume (L)}}_{}](https://img.qammunity.org/2023/formulas/chemistry/college/s234sjp8ee93sr6bgc76xm3o3suykzsymk.png)
Thus, applying the values provided:
![\text{molar concentration of HCl= }\frac{0.01\text{ mol}}{2.5\text{ L}}=0.004\text{ mol/L}](https://img.qammunity.org/2023/formulas/chemistry/college/1m351d4cg5q0g0c54dw9iy7g3pjzjkawex.png)
Therefore, the concentration of the HCl solution is 0.004 mol/L.
Since HCl is a strong acid, we can expect it to completely dissociate in H+ and Cl-: the concentration of H+ ions for this solution would also be 0.004 mol/L ( [H+] = 0.004 mol/L).
III) The next step is calculate the pH of this solution. We'll use the following equation and apply the concentration of H+ ions obtained in the previous step ( [) using H+] = 0.004 mol/L):
![pH=-\log _(10)\lbrack H^+\rbrack\to pH=-\log _(10)(0.004)=2.39](https://img.qammunity.org/2023/formulas/chemistry/college/tc6wvim3iqx05a6m2worntiajzj4ixebn0.png)
Now, we know that the pH of the solution is 2.39.
III) The last step is use the pH value calculated (pH = 2.39) to calculate the pOH of the solution, using the following equation:
![pH+\text{pOH =14}\to pOH\text{ = 14 - pH}](https://img.qammunity.org/2023/formulas/chemistry/college/tv7gej491odxso3qyiheujhwjw5i9kzjnj.png)
![\text{pOH = 14-2.39 = 11.6}](https://img.qammunity.org/2023/formulas/chemistry/college/5a11w7egt8lopmlrgcc6q2p4ajbr0p2cnj.png)
Therefore, the pOH of the HCl solution given is 11.6.