CH 302 1st Edition Lecture 14 Outline of Last Lecture I. Clicker QuestionsII. pH and pOH calculationsOutline of Current Lecture I. Clicker QuestionsII. Neutralization ReactionsIII. Common Ion EffectIV. BuffersCurrent LectureClicker Questions:1. The pH of a solution of a soluble salt will be:A. NeutralB. BasicC. AcidicD. Any of the above, depends on the salt2. Calculate the pH of the following solution, 0.15 M HF; Given the Ka = 7.2 x 10^-4.A. 2.0B. 4.0C. 7.0D. 9.0E. 12.03. Calculate the pH of the following solution, 0.15 M NaF, given the Ka of HF = Ka = 7.2 x 10-4A. 2.0B. 5.9C. 7.0D. 8.1E. 12.0Neutralization Reactions:- What are the components in the following situations?o CH3COOH(aq) + NaOH(aq)NaCH3COO(aq) +H2O(l)o Weak acid + Strong Base1. 100 mL 0.1 M CH3COOH + 100 mL 0.1 M NaOHo CH3COOH + Na+ + OH- (reactants)o CH3COO- + Na+ (products)o Weak base is formed from equal amounts 2. 100 mL 0.1 M CH3COOH + 200 mL 0.1 M NaOHo CH3COOH + Na+ + OH- (reactants)o Na+ + OH- + CH3COO- (products)o A weak base is formedo What dominates pH? OH-!3. 200 mL 0.1 M CH3COOH + 100 mL 0.1 M NaOHo CH3COOH + Na+ + OH- (reactants)o CH3COOH + Na+ + CH3COO- (products)The pH of a 0.1 M aqueous solutions of the salts NaCH3COO, NH4Cl, KCl will be:A) Neutral, Neutral, Neutral B) Basic, Acidic, Neutral C) Acidic, Neutral, Basic D) Basic, Neutral, Acidic E) Acidic, Basic, NeutralCommon Ion Effect:- Percent of ionization = amount ionized divided by initial amount multiplied by 100- The percent of ionization is suppressed in the presence of a common ion- Example:o Percent ionization = 0.00135 M/ 0.1 M x 100% = 1.35%Buffers:- Because the pH changed very little, it is called a buffer solution- Buffer: A solution in which the pH resists change when a strong acid or strong base is added- Buffers can be acidic- Buffers can be basic- Depends on the conjugate acid base pair- What happens if we keep adding NaOH to the solution…o Eventually neutralizes all the acido Excess NaOH with weak baseo Exhausts the buffer- When the initial acid and base are similar in concentration, then the pH is close to the pKa- For the pH to be 1 unit different than the pKa the difference in concentrations must be atleast 10x!- pOH = pKb + log [BH+]/[B]The pKa of HF is 3.18. What is the pH of solution of 100 mL of 0.1 M HF and 100 mL of a 0.2 M NaF?A. slightly less than 3.18 B. 3.18 C. slightly more than 3.18*This is because more base is presentWhat did we learn today?- Weak acids or bases have limited ionization in the presence of a common ion.- Substantial amounts of conjugate acid base pairs, together in solution resist change in pH.- This effect is called buffering. When [HA] = [A-], the pKa = pH of that solution.- When [B] = [BH+], the pKb = pOH of that
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