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 solution, we first need to determine the concentration of hydroxide ions (OH-) in the solution. Since 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 , which is 0.28 M.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.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.Therefore, the pH of the 0.14 M solution is approximately 13.45, and the pOH is approximately 0.55.6. 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.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.The total volume of the solution after adding the acetic acid is 10.05 ml, which is equivalent to 0.01005 L.The concentration of acetic acid in the final solution is given by (0.025 mmol) / (0.01005 L) ≈ 0.0025 M.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:CH3COOH ⇌ H+ + CH3COO-Let the dissociation constant of acetic acid be Ka. The concentration of H+ ions produced by the dissociation is given by:[H+] = √(Ka * [CH3COOH])Substituting the value of Ka for acetic acid (1.8 x 10^-5), we get:[H+] = √((1.8 x 10^-5) * (0.0025)) ≈ 0.00424 MFinally, we calculate the pH using the formula pH = -log[H+]. Substituting the value of [H+], we get pH = -log(0.00424) ≈ 2.37.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.