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Calculate the pOH for an aqueous solution containing 0.01 moles of HCl in a 2.5 L solution.

Sagot :

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:

[tex]\text{molar concentration = }\frac{\text{number of moles (mol)}}{\text{volume (L)}}_{}[/tex]

Thus, applying the values provided:

[tex]\text{molar concentration of HCl= }\frac{0.01\text{ mol}}{2.5\text{ L}}=0.004\text{ mol/L}[/tex]

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):

[tex]pH=-\log _{10}\lbrack H^+\rbrack\to pH=-\log _{10}(0.004)=2.39[/tex]

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:

[tex]pH+\text{pOH =14}\to pOH\text{ = 14 - pH}[/tex][tex]\text{pOH = 14-2.39 = 11.6}[/tex]

Therefore, the pOH of the HCl solution given is 11.6.

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