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Calculating Equilibrium Concentrations in Solutions of Weak Acids

Calculating Equilibrium Concentrations in Solutions of Weak Acids. 14.9. We can us K a to calculate equilibrium concentrations and the pH in a solution of the acid. The key to solving acid-base equilibrium problems is to think about the CHEMISTRY involved

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Calculating Equilibrium Concentrations in Solutions of Weak Acids

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  1. Calculating Equilibrium Concentrations in Solutions of Weak Acids 14.9

  2. We can us Kato calculate equilibrium concentrations and the pH in a solution of the acid. • The key to solving acid-base equilibrium problems is to think about the CHEMISTRY involved • I am referring to the possible proton-transfer reactions that can take place b/w Bronsted-Lowry acids & bases (You’ll see what I mean through examples.)

  3. These equilibrium problems are best solved using a simple 8 step process • Suppose your problem reads: “Calculate the concentrations of all species present (H3O+, CN-, HCN, & OH-) and the pH in a 0.10M HCN solution.”

  4. Step 1 • List the species present & identify them as acids or bases For our current problem it would like this: HCN H2O Acid Acid or Base

  5. Step 2 • Write the possible reactions • Because we have 2 acids (HCN & H2O) and one base (H2O) there are 2 possible reactions • HCN(aq) + H2O(l) --- H3O+(aq) + CN-(aq) Ka=4.9x10^-10 • H2O(l) + H2O(l) ----- H3O+(aq) + OH-(aq) Kw=1.0x10^-14 *Ka value for HCN comes from table 14.2

  6. Step 3 • The reaction that proceeds furthest right (largest K) is called the Principal Reaction • All others are known as Subsidiary Reactions • For our problem Ka for HCN is 10,000 times bigger than Kw • This means the 1st reaction is the Principal

  7. Step 3 (cont’d) • Although the principal reaction and the subsidiary reaction both produce H3O there is only one in the concentration in the solution • We have to assume that almost all of the H3O comes from the stronger acid (HCN) and that the contribution from the weaker acid (H20) is negligible

  8. Step 4 • Set up your ICE table HCN(aq) + H2O(l) --- H3O+(aq) + CN-(aq) Initial 0.10 0 0 Change -x +x +x Equilibrium 0.10-x x x

  9. Step 5 • Set up and solve Ka= 4.9x10^-10= (H30)(CN) = (x)(x) (HCN) (0.10-x) Because Ka is very small the reaction will not proceed very far to the right, making x negligible = (x^2)/0.10

  10. Step 5 (cont’d) x^2 = 4.9x10^-14 X= 7.0x10^-6

  11. Step 6 • Take the number you just found for x and plug it in to the ICE table HCN(aq) + H2O(l) --- H3O+(aq) + CN-(aq) Initial 0.10 0 0 Change -7x10^-6 +7x10^-6 +7x10^-6 Equilibrium 0.10-7x10^-6 7x10^-6 7x10^-6 =0.10

  12. Step 7 • The only concentration unknown for the 2 original reactions is OH • This can be found by using the recently found H3O concentration and Kw Kw= (H3O)(OH) OH= K w =1x10^-14 = 1.4x10^-9 M (H3O) 7x10^-6

  13. Step 8 • Calculate the pH pH= -log (total (H3O)) = -log (7x10^-6) =5.15

  14. Worked Problem 14.10 on page 562 shows you what you would do if your numerical value for x was large enough to not be negligible • Please take a moment to look at it now…YES! NOW! • The thought process to solve the problem is the same, but now you must use the quadratic formula to solve the math

  15. Congratulations! You now know how to calculate the equilibrium concentrations in solutions of weak acids! Try Problem 14.14 a. and b. on page 564 to see if you understand it completely

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