The oxiuc : WClghs 2,02 g culture the p 5. Calculate the pH and pOH of the following strong base solutions :0.14 M Ba(OH)_(2) 6. Calculate the pH of the solution obtained by adding 1 drop (0.05ml) of 0.5 M acetic acid to 10 ml of pure water.

5. To calculate the pH and pOH of the 0.14 M $Ba(OH)_{2}$ solution, we first need to determine the concentration of hydroxide ions (OH-) in the solution. Since $Ba(OH)_{2}$ is a strong base, it dissociates completely in water to produce two hydroxide ions for each formula unit. Therefore, the concentration of OH- ions is twice the concentration of $Ba(OH)_{2}$, which is 0.28 M.<br /><br />Next, we calculate the pOH using the formula pOH = -log[OH-]. Substituting the concentration of OH- ions, we get pOH = -log(0.28) ≈ 0.55.<br /><br />Since pH + pOH = 14 at 25°C, we can calculate the pH using the formula pH = 14 - pOH. Substituting the value of pOH, we get pH = 14 - 0.55 ≈ 13.45.<br /><br />Therefore, the pH of the 0.14 M $Ba(OH)_{2}$ solution is approximately 13.45, and the pOH is approximately 0.55.<br /><br />6. To calculate the pH of the solution obtained by adding 1 drop (0.05 ml) of 0.5 M acetic acid to 10 ml of pure water, we first need to determine the concentration of acetic acid in the final solution.<br /><br />The amount of acetic acid in 0.05 ml of 0.5 M solution is given by (0.05 ml) * (0.5 M) = 0.025 mmol.<br /><br />The total volume of the solution after adding the acetic acid is 10.05 ml, which is equivalent to 0.01005 L.<br /><br />The concentration of acetic acid in the final solution is given by (0.025 mmol) / (0.01005 L) ≈ 0.0025 M.<br /><br />Since acetic acid is a weak acid, it partially dissociates in water to produce hydrogen ions (H+) and acetate ions (CH3COO-). The dissociation can be represented as:<br /><br />CH3COOH ⇌ H+ + CH3COO-<br /><br />Let the dissociation constant of acetic acid be Ka. The concentration of H+ ions produced by the dissociation is given by:<br /><br />[H+] = √(Ka * [CH3COOH])<br /><br />Substituting the value of Ka for acetic acid (1.8 x 10^-5), we get:<br /><br />[H+] = √((1.8 x 10^-5) * (0.0025)) ≈ 0.00424 M<br /><br />Finally, we calculate the pH using the formula pH = -log[H+]. Substituting the value of [H+], we get pH = -log(0.00424) ≈ 2.37.<br /><br />Therefore, the pH of the solution obtained by adding 1 drop (0.05 ml) of 0.5 M acetic acid to 10 ml of pure water is approximately 2.37.