Questions: Calculate the pH and pOH of 0.050 M HCl(aq) Recall: pH = -log [H3O+], pOH = -log [OH-], pH + pOH = 14

Transcript text: Calculate the pH and pOH of $0.050 \mathrm{M} \mathrm{HCl}(\mathrm{aq})$ Recall: $\mathrm{pH}=-\log [\mathrm{H} 3 \mathrm{O}+], \mathrm{pOH}=-\log [\mathrm{OH}-], \mathrm{pH}+\mathrm{pOH}=14$

Solution

Solution Steps

Step 1: Determine the concentration of $\mathrm{H_3O^+}$

Since $\mathrm{HCl}$ is a strong acid, it dissociates completely in water. Therefore, the concentration of $\mathrm{H_3O^+}$ is equal to the concentration of $\mathrm{HCl}$.

\[ [\mathrm{H_3O^+}] = 0.050 \, \mathrm{M} \]

Step 2: Calculate the pH

Using the formula for pH:

\[ \mathrm{pH} = -\log [\mathrm{H_3O^+}] \]

Substitute the concentration of $\mathrm{H_3O^+}$:

\[ \mathrm{pH} = -\log (0.050) \]

Calculate the value:

\[ \mathrm{pH} \approx 1.3010 \]

Step 3: Calculate the pOH

Using the relationship between pH and pOH:

\[ \mathrm{pH} + \mathrm{pOH} = 14 \]

Substitute the value of pH: