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Chapter 6 (CIC) and Chapter 16 (CTCS) Read in CTCS Chapter 16.4-7 Problems in CTCS: 16.23, 25, 27, 31, 33, 35, 37, 41, 45, 57, 49, 51, 59, 61, 63, 65.

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Presentation on theme: "Chapter 6 (CIC) and Chapter 16 (CTCS) Read in CTCS Chapter 16.4-7 Problems in CTCS: 16.23, 25, 27, 31, 33, 35, 37, 41, 45, 57, 49, 51, 59, 61, 63, 65."— Presentation transcript:

1 Chapter 6 (CIC) and Chapter 16 (CTCS) Read in CTCS Chapter 16.4-7 Problems in CTCS: 16.23, 25, 27, 31, 33, 35, 37, 41, 45, 57, 49, 51, 59, 61, 63, 65

2 pH – Power of Hydrogen As [H + ] increases, the pH decreases Based on logarithmic scale Compared to a pH = 1 solution, a pH = 2 solution is 10x less acidic An acid has a pH < 7 A base has a pH > 7

3 pH of Some Common Solutions Brown, T.L.; LeMay, H.E.Jr.; and Bursten, B.E. Chemistry the Central Science, 8 th Edition, Prentice Hall, Upper Saddle River, NJ, 2000, p 602.

4 Why Do We Bother With pH? [H + ] = 3.4 x 10 -6 is too small p stands for –log and H stands for [H + ] pH = -log [H + ] Neutral solution has [H + ] = 1 x 10 -7 or pH = 7.0 Q: If [H + ] = 3.4 x 10 -6, what is the pH? How many sig figs should there be?

5 Q: What is the [OH - ] of this solution? What is the pOH? Notice that pH + pOH = 14.00 (pK w ) Derived from [H + ] [OH - ] = 1 x 10 -14 (K w )

6 Strong Acids/Bases Table 4.2 lists these as: Note that the strong bases are all of the soluble hydroxides Strong Acids HClHBrHIHClO 4 HNO 3 H 2 SO 4 Strong Bases LiOHNaOHKOHRbOH CsOHSr(OH) 2 Ba(OH) 2

7 What’s the definition of a strong acid or base? Q: What is the pH of 0.008 M HCl? Q: The pH of the HCl soln in lab was about 1.85. What was it’s concentration?

8 Q: What is the [OH - ] in 0.010 M Sr(OH) 2 ? Q: What is the pH of this solution? Q: What is the pH of a saturated Ca(OH) 2 solution if the solubility is 0.97 g/100. mL?

9 Weak Acids Not to be confused with dilute acids! 0.0001 M HCl is NOT a weak acid! HClO + H 2 O  H 3 O + + ClO - Remember that K = [products]/[reactants]

10 If [HClO] = 1.0 M, then [H 3 O + ][ClO - ] = 3.0 x 10 -8 So, since there should be equal amounts of [H 3 O + ] and [ClO - ], then [H 3 O + ] 2 = 3.0 x 10 -8 Or [H 3 O + ] = 1.7 x 10 -4 And pH = 3.76

11 Things to Remember A K a must have H + in the products K a ’s of strong acids have large values (greater than 1) A K b must have OH - in the products K b ’s of strong bases have large values (greater than 1) You may not be able to simplify the problem so much and you will have to make approximations

12 Q:Calculate the pH of a 0.025 M lactic acid solution if its’ K a = 1.4 x 10 -4. A: 2.75

13 Q:Calculate the K a of a 0.085 M phenylacetic acid solution if its’ pH is 2.68. A: 5.3 x 10 -5

14 Q:Calculate the pH of a 1.0 M methylamine solution if its’ K b = 4.38 x 10 -4. A: 12.32


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