Chem 116 POGIL Worksheet - Week 9 - Solutions Weak Acid and Base Equilibria

Transcription

1 Chem 116 POGIL Worksheet - Week 9 - Solutions Weak Acid and Base Equilibria Key Questions 1. Given the following concentrations of H3O or OH, calculate the concentration of OH or H O, and indicate whether the solution is acidic or basic. 3 [H3O ] [OH ] Acidic or basic? x 10 M 1.0 x 10 M basic x 10 M 4.0 x 10 M acidic x 10 M 8.3 x 10 M basic 2. Complete the following table by calculating the missing entries and indicate whether the solution is acidic or basic. Be sure your answers are expressed to the proper number of significant figures. [H3O ] [OH ] ph poh acidic or basic? x 10 M 1.7 x 10 M acidic x 10 M 6.2 x 10 M basic x 10 M 1.7 x 10 M basic x 10 M 2.0 x 10 M acidic M 5.0 x 10 M acidic (very!) 3. Determine the concentrations of H3O and OH in the following solutions of strong acids or bases in water. [Caution: Think about how the acid or base dissociates in water.] Solution [H3O ] [OH ] x 10 M HCl 2.5 x 10 M 4.0 x 10 M M NaOH 1.0 x 10 M 1.0 x 10 M x 10 M Ca(OH) x 10 M 1.5 x 10 M In the last case, realize that two moles of OH are produced for every one mole of Ca(OH) 2 2 by the dissociation reaction, Ca(OH) (aq) Ca (aq) 2 OH (aq). 2

2 4. A M solution of a weak acid HA has a ph of What is the value of K a for the acid? [Hint: What is the actual concentration of undissociated HA, [HA], in this solution?] 3 ph = [H3O ] = 2.5 x 10 M We can assume that the acid is the major source of hydronium ion and that water contributes little to the total [H3O ]. This concentration of hydronium ion, then, also represents the amount of the initial HA that is lost through hydrolysis. On this basis we have the following equilibrium concentrations: HA H2O H3O A Initial Equilibrium x x 10 = What is the ph of M benzoic acid (C6H5CO2H = HBz), for which K a = ? Is it necessary to solve the quadratic equation in this case? HBz H2O H3O Bz The concentration of the acid and the K a fit our criteria to avoid needing the quadratic equation. ph = 2.79 The percent dissociation of the benzoic acid, based on our calculation, is This is acceptable, so the quadratic equation was not necessary.

3 3 6. Find the concentrations of all species and ph for a M HF solution. K a = You will need to find values for [HF], [H O ], [F ], and [OH ]. Trying the quick method, we obtain HF H2O H3O F This is not acceptable. Notice that C HA = 1.0 x 10 and K a = 6.7 x 10 are within two powers of 10 of each other, suggesting that HF is appreciably dissociated. We must solve the quadratic. This means that we will assume [HF] = C HF [H3O ]. Substituting into the K expression: a [H3O ] = [H3O ] [H3O ] [H3O ] = 0 3 Solving the quadratic equation and ignoring the negative root ( ) gives 4 [H3O ] = M = [F ] ph = [HF] = C HF [H3O ] = = M To find [OH ], use K w:

4 5 7. What is the ph of a M solution of NH 3(aq)? For ammonia, K b = Is it necessary to solve the quadratic equation? Justify the approach you took, either way. The hydrolysis equilibrium and the equilibrium concentrations are as follows: NH 3 H2O NH 4 OH x x x The analytical concentration of ammonia is M, which is greater than K b = by much more than two powers of 10. Therefore, we can probably ignore the loss of NH 3 from hydrolysis. Thus, we can write and From this result, the percent hydrolysis is much less than 5%: Therefore, we are justified in ignoring loss of NH 3 from hydrolysis and avoiding the quadratic equation. To obtain ph, we could calculate [H3O ] = K w/[oh ], and then get ph; or we could calculate poh from our value of [OH ], and get ph from ph = poh. Taking the second approach, we have 3 poh = log( ) = ph = = Sodium hypochlorite, NaOCl, is the active ingredient in chlorine bleach (e.g., Chlorox ). 8 For hypochlorous acid, HOCl, K = a a. Write the equation for the base hydrolysis equilibrium of the hypochlorite ion. OCl H2O HOCl OH b. What is the value of K b for the hypochlorite ion?

5 c. What is the ph of a 0.10 M solution of sodium hypochlorite? In solution, the 0.10 M NaOCl will break up into 0.10 M Na and 0.10 M OCl. Therefore, the analytical concentration of the base is. NaOCl Na OCl OCl H2O HOCl OH 0.10 x x x Because the analytical concentration of the base is so much greater than K b, loss of OCl is negligible, and we can assume. Therefore, From this, poh = 3.74, and ph = = Trimethylamine, (CH 3) 3N, is a weak base (K b = ) that hydrolyzes by the following equilibrium: (CH 3) 3N H2O (CH 3) 3NH OH a. Write the equation for the acid hydrolysis equilibrium of trimethylammonium ion, (CH ) NH. 3 3 (CH 3) 3NH H2O (CH 3) 3N H3O b. What is the value of K a for the trimethylammonium ion, (CH 3) 3NH?

6 c. What is the ph of a 0.12 M solution of trimethylammonium chloride, ((CH 3) 3NH)Cl? ((CH 3) 3NH)Cl (CH 3) 3NH Cl (CH 3) 3NH H2O (CH 3) 3N H3O x x x 10 a With K = , (CH ) NH is a very weak acid, so we can definitely ignore hydrolysis and assume [(CH ) NH ] =. 10. Explain the following observations: ph = 5.36 a. H2Se is a stronger acid than H2S Se is larger than S, so the SeH bond is weaker, making H Se a stronger acid. b. HNO 2 is a weak acid, but HNO 3 is a strong acid HNO has one terminal oxygen, but HNO has two, making it a stronger acid. 2 3 c. H3PO 4 is a stronger acid than H3AsO4 P is more electronegative than As, making H PO a stronger acid. d. FCH2CO2H is a stronger acid than CH3CO2H Electronegative fluorine withdraws electron density from the -OH bond of the carboxyl group, -CO H, making FCH CO H a stronger acid