Acids Lesson 8 Application of Hydrolysis pH Calculations.

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Presentation transcript:

Acids Lesson 8 Application of Hydrolysis pH Calculations

pHEquations You must know the following equations, which are all based on the ionization of water at 25 0 C! H 2 O ⇄ H + +OH - Kw=[H + ][ OH - ]=1.00 x pH=-Log[H + ]pOH=-Log[OH - ] [H + ]=10 -pH [OH - ]=10 -pOH pH+pOH=pKw=14.000

OperationInverse - Log 10 x x ÷ Sig Figs and pH 1.00 x M 3 sig figs -Log 1.00 x = sig figs The 4 is from the which does not count for sig figs For pH or POH only digits after the decimal are significant +

pHCalculations [H + ][OH - ]pHpOH 2.5 x M Calculation

pHCalculations [H + ][OH - ]pHpOH 2.5 x M CalculationThe number of sig figs are 2 (green)

pHCalculations [H + ][OH - ]pHpOH 2.5 x M CalculationpH = -Log[H + ] pH = -Log[2.5 x ] pH = 3.60 The digit before the decimal does not count as a sig fig. Both digits after the decimal count for 2 sig figs.

pHCalculations [H + ][OH - ]pHpOH 2.5 x M3.60 CalculationpH = -Log[H + ] pH = -Log[2.5 x ] pH = 3.60 keep all digits on the calculator- do not round!

pHCalculations [H + ][OH - ]pHpOH 2.5 x M3.60 CalculationpH + pOH = pOH = pOH = 10.40

pHCalculations [H + ][OH - ]pHpOH 2.5 x M Calculation[OH - ]=10 -pOH [OH - ]= [OH - ]=4.0 x M

pHCalculations [H + ][OH - ]pHpOH 2.5 x M4.0 x M CalculationNote that all of the entries in the last line have two sig figs!

pHCalculations [H + ][OH - ]pHpOH 2.5 x M4.0 x M CalculationpH + pOH = pOH = pOH = 4.565

pHCalculations [H + ][OH - ]pHpOH 2.5 x M4.0 x M Calculation[OH - ]=10 -pOH [OH - ]= [OH - ]=2.72 x M

pHCalculations [H + ][OH - ]pHpOH 2.5 x M4.0 x M x M Calculation[H + ][OH - ]=1.00 x [H + ][2.72 x ]=1.00 x [H + ]=3.67 x M

pHCalculations [H + ][OH - ]pHpOH 2.5 x M4.0 x M x M2.72 x M CalculationNote that all of the entries in the last line have three sig figs!

pHCalculations [H + ][OH - ]pHpOH 2.5 x M4.0 x M x M2.72 x M x M CalculationpOH = -Log[OH - ] pOH = -Log[2.6 x ] pOH = 3.59

pHCalculations [H + ][OH - ]pHpOH 2.5 x M4.0 x M x M2.72 x M x M3.59 CalculationpH + pOH = pH = pOH = 10.41

pHCalculations [H + ][OH - ]pHpOH 2.5 x M4.0 x M x M2.72 x M x M Calculation[H + ]=10 -pH [H + ]= [H + ]=3.8 x M

pHCalculations [H + ][OH - ]pHpOH 2.5 x M4.0 x M x M2.72 x M x M 2.6 x M CalculationNote that all of the entries in the last line have two sig figs!

pHCalculations [H + ][OH - ]pHpOH 2.5 x M4.0 x M x M2.72 x M x M 2.6 x M

pHCalculations [H + ][OH - ]pHpOH 2.5 x M4.0 x M x M2.72 x M x M 2.6 x M x M8.40

pHCalculations [H + ][OH - ]pHpOH 2.5 x M4.0 x M x M2.72 x M x M 2.6 x M x M8.40

pHCalculations [H + ][OH - ]pHpOH 2.5 x M4.0 x M x M2.72 x M x M 2.6 x M x M4.0 x M8.40

pHCalculations [H + ][OH - ]pHpOH 2.5 x M4.0 x M x M2.72 x M x M 2.6 x M x M4.0 x M

pHCalculations [H + ][OH - ]pHpOH 2.5 x M4.0 x M x M2.72 x M x M 2.6 x M x M4.0 x M CalculationNote that all of the entries in the last line have two sig figs!

pHCalculations [H + ][OH - ]pHpOH 2.5 x M4.0 x M acid 3.67 x M2.72 x M basic 3.8 x M 2.6 x M basic 2.5 x M4.0 x M acid If the pH is lower than 7, the solution is acidic! If the pH is greater than 7, the solution is basic!

1.Calculate the pH of 0.40 M HI. HI  H + +I M0.40 M0.40 M pH = -Log[H + ] pH = -Log[0.40] pH = 0.40

2.Calculate the pH of M Ba(OH) 2. Ba(OH) 2  Ba 2+ +2OH M0.030 M0.060 M pOH = -Log[OH - ] pOH = -Log[0.060] pOH = 1.22 pH + pOH = pH = pH = 12.78

3.Calculate the pH of mL of 1.0 M HCl after mL of water is added to it. HCl  H + +Cl M M0.25 M pH=0.60