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PH. The Ion product constant for water H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH − (aq) Dissociation of water is an equilibrium, and since liquid water.

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Presentation on theme: "PH. The Ion product constant for water H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH − (aq) Dissociation of water is an equilibrium, and since liquid water."— Presentation transcript:

1 pH

2 The Ion product constant for water H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH − (aq) Dissociation of water is an equilibrium, and since liquid water has no effect, we can calculate the ion product constant for water: Kw = [H 3 O + ][OH − ]

3 The Ion product constant for water Very few (2/billion) water molecules dissociate. The [H3O + ] and [OH − ] were determined to be 1.0 x 10 -7 mol/L each. Kw =1.0×10 −7 mol/L × 1.0×10 −7 mol/L = 1.0 ×10 −14

4 The Ion product constant for water Since we know that at equilibrium the K for water (Kw) is 1.0 x 10 -14 we can determine the [H3O + ] and [OH − ] if one is known.

5 Problem Find [H3O + ] and [OH − ] in each solution. (a) 2.5 mol/L nitric acid (b) 0.16 mol/L barium hydroxide You know that nitric acid is a strong acid and barium hydroxide is a strong base. Since both dissociate completely in aqueous solutions, you can use their molar concentrations to determine [H3O+] or [OH−]. You can find the concentration of the other ion using Kw.

6 Practice Problems P 566 #12-15

7 The pH Scale Strong acids contain 10 mol/L of H 3 O + Strong bases contain 10 -15 mol/L of H 3 O + Since these numbers are inconvenient we use the pH scale to simplify. Logarithms are used to calculate pH

8 Logarithms The log of any number is the power to which you must raise 10 to equal that number: log 10 10=1 because 10 1 =10 log 10 100=2 because 10 2 =100 Log 10 1000=3 because 10 3 =1000

9 Calculating pH pH= - log[H 3 O + ] Sometimes [H + ] is used as an abbreviation. p stands for exponential “power” of 10 [H 3 O + ] or [H + ] is the concentration of acid in mol/L

10 Calculating pH

11 Problem Calculate the pH of a solution with [H 3 O + ] = 3.8 × 10 −3 mol/L.

12 Practice P 569 16-19

13 Calculating pOH pOH refers to the ion concentration of hydroxide. Calculate pOH using the [OH - ] pOH= - log[OH - ] Since Kw= [H3O + ][OH − ]= 1.0x10 -14 Then, pH + pOH= 14

14 Problem A liquid shampoo has a hydroxide ion concentration of 6.8 ×10 −5 mol/L at 25°C. (a) Is the shampoo acidic, basic, or neutral? (b) Calculate the hydronium ion concentration. (c) What is the pH and the pOH of the shampoo?

15 Using antilogs You can calculate [H 3 O + ] or [OH − ] by finding the antilog of the pH or pOH. [H 3 O + ] = 10 −pH [OH − ] = 10 −pOH To use the antilog, push 2 nd and then log on your calculator. It should give you 10 ^( then insert –pH or –pOH

16 Problem If the pH of urine is outside the normal range of values, this can indicate medical problems. Suppose that the pH of a urine sample was measured to be 5.53 at 25°C. Calculate pOH, [H3O+], and [OH−] for the sample.

17 Practice P 572 # 20-25 Section Review: P 578 # 1-5, and 8, especially 8 Quiz Friday May 2 nd Acid/Base Properties pH/pOH [H 3 O + ] / [OH - ]


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