A solution of acetic acid is 2.0% dissociated at 25.0°C. What was the original concentration (in M) of the acetic acid solution? The Ka at 25.0 °C for acetic acid is 1.8 × 10^-5.
A solution of acetic acid is 2.0% dissociated at 25.0°C. What was the original concentration (in M) of the acetic acid solution? The Ka at 25.0 °C for acetic acid is 1.8 × 10^-5.
General Chemistry - Standalone book (MindTap Course List)
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Chapter16: Acid-base Equilibria
Section: Chapter Questions
Problem 16.112QP: Ionization of the first proton from H2SeO4 is complete (H2SeO4 is a strong acid); the...
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Question
- A solution of acetic acid is 2.0% dissociated at 25.0°C. What was the original concentration (in M) of the acetic acid solution? The Ka at 25.0 °C for acetic acid is 1.8 × 10^-5.
- A solution of ammonia is 3.2% ionized at 25.0°C. What was the original concentration (in M) of the ammonia solution? The Kb at 25.0°C for ammonia is 1.8×10^-5.
- Calculate the pOH of 0.716 M anilinium hydrochloride (C6H5NH3Cl) solution in water, given that Kb for aniline is 3.83 × 10^-4.
- The acid-dissociation constant of hydrocyanic acid (HCN) at 25.0°C is 4.9 × 10^-10. What is the pH of an aqueous solution of 0.080 M sodium cyanide (NaCN)?
- The base-dissociation constant of ethylamine (C2H5NH2) is 6.4 × 10^-4 at 25.0 °C. The [H+] in a 1.6 × 10^-2 M solution of ethylamine is __________ M.
- The acid-dissociation constants of phosphoric acid (H2PO3) are Kal = 7.5 × 10^-3, Ka2 = 6.2 × 10^-8, and Ka3 = 4.2 × 10^-13 at 25.0°C. What is the pH of a 2.5 M aqueous solution of phosphoric acid?
- What is the pH of a sodium formate solution prepared by adding 0.680 grams of sodium formate to 100.0 ml of water at 25.0°C? The Ka at 25.0°C for formic acid is 1.8 × 10^-4.
- Calculate the pH of a buffer that contains 0.270 M hydrofluoric acid (HF) and 0.180 M cesium fluoride (CsF). The Ka of hydrofluoric acid is 6.80 × 10^-4.
- The Kb of ammonia is 1.76 × 10^-5. The pH of a buffer prepared by combining 45.0 mL of 0.180 M ammonia and 40.0 mL of 0.205 M ammonium nitrate is __________.
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