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K a

K a. What is K a ?. No, it isn’t Ka. Recall…. …we assume that both the strong acid and strong base completely dissociate (break apart into ions) when dissolved in water. Thus, we assume in a strong acid and a strong base, the neutralization reaction results in everything as an ion.

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K a

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  1. Ka

  2. What is Ka? • No, it isn’t Ka.

  3. Recall…. …we assume that both the strong acid and strong base completely dissociate (break apart into ions) when dissolved in water. Thus, we assume in a strong acid and a strong base, the neutralization reaction results in everything as an ion. For example: HCl(aq) + NaOH(aq) HOH + NaCl(aq) Is really H+ + Cl- + Na+1 + OH-  H+ + OH- + Na+ + Cl- Thus, it is easy to calculate pH or pOH. Remember pH = -log [H+] and pOH = -log [OH-].

  4. But what happens if you have a weak acid or base? A weak acid or a weak base does NOT completely dissociate. Thus, you can’t really measure the total number of H+ or OH- in the reaction. Thus, no real pH or pOH. So now what??????

  5. This is where Ka comes in Ka is a measure of the extent to which a weak acid really ionizes. Ka is called the acid-dissociation constant. It is similar to the equilibrium constant Keq

  6. For Example CH3COOH + H2O  H3O+ + CH3COO- Ka= [H3O+][CH3COO-] [CH3COOH] This Ka value now tells us the tendency of CH3COOH to ionize in water.

  7. Ka Values A high Ka value = the stronger the acid (more ionization) Example: Ka of HF = 6.8 x 10-4 A small Ka value = the weaker the acid (less ionization) Example: Ka of C6H5O- = 1.3 x 10-10

  8. What do we use Ka values for? Since weak acids only partially dissociate, the calculation of pH is suspicious. Thus, Ka is used to determine pH of weak acids. You can also use pH of weak acids to calculate Ka.

  9. Calculating Ka from pH A student prepared a 0.10M solution of HCOOH and measure the pH. The pH was 2.38. Calculate the Ka for HCOOH. HCOOH H+ + HCOO- Ka = [H+][HCOO-] [HCOOH] pH = 2. 38 Thus [H+]=10-2.38 = 4.2 x 10-3

  10. More Example #1 HCOOH H+ + HCOO- Ka = [H+][HCOO-] = (4.2x10-3)(4.2x10-3) = 1.8x10-4 [HCOOH] 0.10-4.2 x10-3

  11. Ka from pH—Example #2 A student prepared a 0.020M solution of HNO2 and measured the pH. The pH was 3.26. Calculate the Ka for HNO2. HNO2 H+ + NO2- Ka = [H+][NO2-] [HNO2] pH = 3.26 Thus [H+]=10-3.26 =0.0005495 M

  12. Ka from pH—Example #2 continued HNO2 H+ + NO2- Ka = [H+][NO2-] = (0.0005495)(0.0005495) = 0.000155 [HNO2] 0.020–0.0005495

  13. Even More Ka from pH Now watch this video for Example #3: Find Ka from pH - Example #3 Do #1 & 3 on Brown worksheet. Check my website for answers.

  14. Calculating pH from Ka – Example 1 Calculate the pH of a 0.30 M solution of CH3COOH given Ka=1.8 x 10-5 CH3COOH H+ + CH3COO- Ka = [H+][CH3COO-] = 1.8 x 10-5 [CH3COOH]

  15. More pH from Ka – Example 1 Ka = [H+][CH3COO-] = (x)(x) = 1.8 x 10-5 [CH3COOH] 0.30-x The x is so small in comparison to the Molarity of CH3COOH to begin with, we assume Molarity – x is negligible and just drop the x from the formula. (x)(x) = 1.8 x 10-5 0.30

  16. Even More pH from Ka – Example 1 (x)(x) = 1.8 x 10-5 0.30 x2 = (1.8 x 10-5)(0.30) = 5.4 x 10-6 x = 2.3 x 10-3 x = [H+] pH = -log[H+] pH = -log[2.3 x 10-3] = 2.64

  17. Calculating pH from Ka – Example 2 Calculate the pH of a 0.20 M solution of HCN given Ka= 4.9 x 10-10 HCN H+ + CN- Ka = [H+][CN-] = 4.9 x 10-10 [HCN]

  18. More pH from Ka – Example 2 Ka = [H+][CN-] = (x)(x) = 4.9 x 10-10 [HCN] 0.20-x The x is so small in comparison to the Molarity of HCN to begin with, we assume Molarity – x is negligible and just drop the x from the formula. (x)(x) = 4.9 x 10-10 0.20

  19. Even More pH from Ka – Example 2 (x)(x) = 4.9 x 10-10 0.20 x2 = (4.9 x 10-10)(0.20) = 9.8 x 10-11 x = 9.8 x 10-6 x = [H+] pH = -log[H+] pH = -log[9.6 x 10-6] = 5.00

  20. Even More pH from Ka Now Watch this Video for one more example. Find pH from Ka - Example 3 Do #4 on the Brown worksheet. Check my website for answers.

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