S11-HW4-CH16

# S11-HW4-CH16 - HOMEWORK WEEK 4 CHAPTER 16 4.27 b First...

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HOMEWORK WEEK 4 CHAPTER 16 4.27 b. First, determine the concentrations of the compounds after mixing together. Use the dilution relationship, M 1 V 1 = M 2 V 2 . Since equal volumes of equal molar solutions are mixed, the resulting concentrations are 0.50 M KBr and 0.50 M K 3 PO 4 .There is one Br - ion per KBr, so the concentration of Br - is 0.50 M. There is also one PO 4 3- ion per K 3 PO 4 , so its concentration is also 0.50 M. Potassium ion can be determined as follows. 0.50 M KBr x + 1 mol K 1 mol KBr + 0.50 M K 3 PO 4 x + 3 4 3 mol K 1 mol K PO = 2.0 M K + 010. The reaction with labels of “acid” and “base” written below is as follows: acid base OH - ( aq ) HF( aq ) F - ( aq ) H 2 O( l ) + + base acid We can construct this equation by adding the following two equations: HF(aq) + H 2 O(l) F - (aq) + H 3 O + (aq), K 1 = 6.8 x 10 -4 (Table 16.1) H 3 O + (aq) + OH - (aq) H 2 O(l) + H 2 O(l), K 2 = 1.0 x 10 14 (= 1/K w ) Thus, K c for the given reaction is K 1 x K 2 = 6.8 x 10 10 . Note that this is a very large number, as expected for a reaction of a strong base. ----- We can also apply our equation: K c = a b w K of acid-L x K of base-L K K a of HF is 6.8 x 10 -4 (see K 1 above) K b of OH - is 1 (see Acid-Base Pairs list) Thus K c = 6.8 x 10 -4 / 1.0 x 10 -14 = 6.8 x 10 10 , as above. ----- For the reverse reaction, F - (aq) + H 2 O(l) HF(aq) + OH - (aq) K c is the inverse of the above value, i.e. 1/6.8 x 10 10 = 1.5 x 10 -11 . Note that this reaction is the hydrolysis of the F - ion, and its K c value is equal to K b of F - , i.e. to K w /K a of HF, using our rule that K a x K b = K w for a conjugate pair. 16.3 Both methods involve direct measurement of the concentrations of the hydronium ion and the anion of the weak acid and calculation of the concentration of the un-ionized acid. All concentrations are substituted into the K a expression to obtain a value for K a . In the first method, the electrical conductivity of a solution of the weak acid is measured. The conductivity is proportional to the concentration of the hydronium ion and anion. In the second method, the pH of a known starting concentration of weak acid is measured. The pH is converted to [H 3 O + ], which will be equal to the [anion]. The second method is analogous to a Type 2 problem.

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16.12 {OCL} approach : we can say that CH 3 NH 3 Cl is acidic (the active species, CH 3 NH 3 + , is the conjugate acid of CH 3 NH 2 ), so the pH MUST decrease when this acidic species is added. The Solutions Manual says: The addition of CH 3 NH 3 Cl to 0.10 M CH 3 NH 2 exerts a common-ion effect that causes the following equilibrium to exhibit a shift in composition to the left: CH 3 NH 2 (aq) + H 2 O(l) CH 3 NH 3 + (aq) + OH - (aq) This shift lowers the equilibrium concentration of the OH - ion, which increases the [H 3 O + ]. An increase in [H 3 O + ] lowers the pH below 11.8. The shift in composition to the left occurs according to Le Chatelier's principle, which states that a system
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S11-HW4-CH16 - HOMEWORK WEEK 4 CHAPTER 16 4.27 b First...

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