CH 102 1st Edition Lecture 14Outline of Current Lecture I. Example ProblemsII. Autoionizaton of waterIII. A solution of hydrogen fluoride in waterIV. A solution of sodium fluoride in waterV. Step By Step ProblemVI. What happens?Current LectureI. Example Problemsa. What is the pH of a 0.210 M solution of ammonia in water?i. 11.2 **b. If a ammonia is dissolved in water to make a 0.210 M solution, what is the concentration of hydroxide ions in the solution?i. 0.00191 **c. What is the pH of an aqueous solution containing 0.456 M nitric acid?i. 0.341 **d. What is the concentration of hydronium ions in an aqueous solution of 0.456 M nitric acid?i. 0.456 **These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.II. Autoionizaton of watera. H2O (aq) + H2O (aq) → H3O+ (aq) + OH- (aq) b. At neutral:i. [H3O+] = [OH-] = 1.0 x 10-7ii. pH = pOH = 7III. A solution of hydrogen fluoride in watera. HF (aq) + H2O (aq) → H3O+ (aq) + F- (aq) b. In the forward reactioni. Hydrogen fluoride is acting as an acid. It donates a proton to a water molecule.ii. The water molecule acts as a base. It accepts a proton from a hydrogen fluoride molecule.iii. The product hydronium ion is the conjugate acid.iv. The product fluoride ion is a conjugate base.c. In the reverse reactioni. The hydronium ion acts as an acid and donates a proton to the fluoride ionii. The fluoride ion acts as a base and accepts a proton from the hydronium ionIV. A solution of sodium fluoride in watera. NaF (aq) + H2O (aq) → HF (aq) + Na+ (aq) + OH- (aq) b. In the forward reactioni. Fluoride ion is acting as a base. It accepts a proton from a water molecule.ii. The water molecular acts as an acid. It donates a proton to a fluoride ion.iii. The product hydrogen fluoride is the conjugate acid.iv. The product hydroxide ion is a conjugate base.c. In the reverse reactioni. The hydrogen fluoride molecule acts as an acid and donates a proton to the hydroxide ionii. The hydroxide ion acts as a base and accepts a proton from the hydrogen fluoride moleculeV. Step By Step:a. What is the pH of a 1.35 x 10-3 M solution of sodium fluoride in water?i. Since the pKb is so small you can use the simplifying assumptionii. Solve the problem in steps:1. Determine the hydroxide ion concentration2. Determine the hydronium ion concentration3. Determine the pHiii. Alternative1. Determine the hydroxide ion concentration2. Determine the pOH3. Determine the pHVI. What happens?a. When you add an acid to water i. Hydronium ions are formedii. the pH is low and the solution is acidicb. When you add the conjugate base of an acid to wateri. Hydroxide ions are formedii. the pH is high and the solution is basicc. When you add a base to wateri. Hydroxide ions are formedii. The pH is high and the solution is basicd. When you add the conjugate acid of a base to wateri. Hydronium ions are formedii. The pH is low and the solution is
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