Approximately how much water should be added to 10.0 mL of 11.1 M HCl so that it has the same pH as 0.90 M acetic acid (K_a = 1.8 $\times$ 10^-⁵)? A)28 mL B)276 mL C)3 L D)28 L E)276 L

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Quizplus · Accepted Answer

1. Calculate the pH of 0.90 M acetic acid: K_a = 1.8 F1F1F1S1F1F1F10 10^-⁵ - Convert the Ka value to a pKa value: pKa = -log(Ka) = -log(1.8 x 10^-5) = 4.74 - Use the Henderson-Hasselbalch equation to calculate the pH: pH = pKa + log([A-]/[HA]) - [A-] = concentration of acetate ion = 0.90 M - [HA] = concentration of acetic acid = 1.0 M pH = 4.74 + log(0.90/1.0) = 4.63 2. Calculate the initial pH of 10.0 mL of 11.1 M HCl: - HCl is a strong acid, so it completely dissociates in water: HCl → H+ + Cl- - Calculate the concentration of H+ ions: [H+] = 11.1 M - Calculate the pH: pH = -log[H+] = -log(11.1) = 1.05 3. Calculate the volume of water needed to dilute the HCl solution: - Let x be the volume of water added (in mL) - The final concentration of H+ ions should be the same as in the acetic acid solution: [H+] = 0.90 M - Use the dilution equation: M1V1 = M2V2 - M1 = initial concentration = 11.1 M - V1 = initial volume = 10.0 mL - M2 = final concentration = 0.90 M - V2 = final volume = 10.0 mL + x Substituting the values: (11.1)(10.0) = (0.90)(10.0 + x) x = 275.5 mL Therefore, the answer is D) 28 L (275.5 mL = 0.2755 L, which is approximately 28 L when rounded to two significant figures).

Approximately How Much Water Should Be Added to 10 ×\times× 10-5)? A)28 ML B)276 ML C)3 L D)28 L E)276

Approximately How Much Water Should Be Added to 10 $\times$ 10^-⁵)?
A)28 ML
B)276 ML
C)3 L
D)28 L
E)276