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- pH and pOH 1.  pH scale developed by a Danish chemist Sören Sörenson  Numerical value without units, that communicates the hydrogen ion concentration.

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Presentation on theme: "- pH and pOH 1.  pH scale developed by a Danish chemist Sören Sörenson  Numerical value without units, that communicates the hydrogen ion concentration."— Presentation transcript:

1 - pH and pOH 1

2  pH scale developed by a Danish chemist Sören Sörenson  Numerical value without units, that communicates the hydrogen ion concentration of a solution  The pH of a solution is the negative of the logarithm to the base ten of the hydrogen ion concentration pH = -log[H + (aq) ] 2

3  The number of significant digits following the decimal point in the pH value is equal to the number of significant digits in the hydrogen ion concentration Example Calculate the pH of a solution with a hydrogen ion concentration of 4.57 x mol/L. Answer 4.57 x mol/L has three significant digits, therefore pH = has three decimal places 3

4  At SATP,  an acidic solution is one in which the [H + (aq) ] is GREATER than mol/L  a basic solution is one where the [H + (aq) ] is LESS than mol/L  A neutral solution is one where the [H + (aq) ] is EQUAL to mol/L 4

5  Neutral solutionpH = 7.00  Acidic solutionpH < 7.00  Basic solutionpH > 7.00  NOTE that the hydrogen ion concentration changes by a multiple of 10 for every increase or decrease in one pH unit. 5

6  Sometimes converting from pH to the molar concentration of hydrogen may be necessary  Conversion based on the mathematical concept that a base ten logarithm represents an exponent [H + (aq) ] = 10 -pH Example Convert a pH of to a hydrogen ion concentration. Answer pH = has two decimal places, therefore the hydrogen ion concentration is 4.4 x mol/L with two significant digits. 6

7  It is convenient to describe hydroxide ion concentrations in a similar way as is done for H + (aq) concentrations, by calculating pOH pOH = -log[OH - (aq) ]  A solution’s pOH may be used to calculate the hydroxide ion concentration: [OH - (aq) ] = 10 -pOH 7

8 p Kw = -logK w  Since we know Kw = 1.0 x at SATP the –logK w = = pK w  Using logarithm rules and equilibrium law expression for the autoionization of water we can say pH + pOH = pK w pH + pOH = (at SATP) 8

9 p. 546 Practice UC # 12, 13, 14, 15(a)(b)(c) p. 549 Practice UC # 17, 18, 19 p. 549 Section 8.1 Questions UC # 4 9


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