pH calculations

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pH calculations

• strong acids/strong bases

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pH scale

• Below 7 is acidic
• Above 7 is basic or alkaline
• 7 is neutral

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• The controlling equilibrium in water solutions
  is H2O H2O ?? H3O OH-
• H3O OH- 1 x10-14
• If an acid is added to water, the H3O
  increases while the OH- decreases. Their
  product stays the same (1 x10-14) .
• If a base is added to water, the OH- increases
  while the H3O decreases. Their product stays
  the same (1 x10-14).

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Finding the pH if H3O is exactly 10 (whole
number)

• Find the pH of a .01M HCl solution
• Since the HCl is a strong acid, it dissociates
  100 HCl H2O ? H3O Cl- H3O .01M
• .01M 1 x 10-2
• pH -log H3O
• pH - log (10-2)
• remember logarithms are exponents .. So the pH
  -(-2) 2

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Example 1

• Find the pH of a .001M solution of HNO3
• H3O .001 mol/L 10-3M
• pH -logH3O
• pH - log(10-3)
• pH -(-3)
• pH 3

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Finding the pH if H3O is not exactly 10
(whole number)

• Find the pH of a .075M solution of HBr
• HBr is a strong acid so the H3O .075M
• pH - log (.075) -log (7.5 x 10-2)
• pH -(-1.12) 1.12

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Example 2

• Find the pH of a .005M solution of HClO4.
• HClO4 is a strong acid so H3O .005M
• pH - log (.005)
• pH -(-2.30) 2.30

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Finding the pH if OH- is exactly 10 (whole
number)

• pOH is frequently used with bases.
• pOH - log OH-
• pH pOH 14 for all water solutions
• Find the pH of a .1M solution of NaOH
• NaOH is a strong base so the molarity of the
  baseOH-.
• pOH - log (10-1)
• pOH -(-1)
• pOH 1
• pH 14- pOH 13

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Example 3

• Find the pH of a .0001M solution of KOH
• Since KOH is a strong base, OH- .0001M
• pOH - log (.0001)
• pOH -(-4) 4.0
• pH14- 4 10.0

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Finding the pH if OH- is not exactly 10
(whole number)

• Find the pH of a .065M solution of LiOH
• Since LiOH is a strong base, OH- .065M
• pOH - log(.065)
• pOH -(-1.19) 1.19
• pH 14 1.19 12.81

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Example 4

• Find the pH of a .00325M solution of CsOH
• Since CsOH is a strong base, OH- .00325M
• pOH - log(.00325)
• pOH -(-2.49) 2.49
• pH 14-2.49 11.51

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Polyprotic acids or polyhydroxic bases.

• Strong acids such as H2SO4 will dissolve in water
  and produce 2 moles of H3O for one mole of
  acid.
• So the H3O of H2SO4 is twice the acid molarity
• The pH of a .05M H2SO4 solution is .
• pH - log (.1)
• pH -(-1) 1
• In a similar fashion, the OH- of a .02M Ba(OH)2
  solution is twice the base molarity and is
  treated the same way as for a polyprotic acid.
• pOH -log(.04) 1.39
• pH 14-1.39 12.61

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pH to H3O calculations

• H3O inv log(-pH) 10-pH
• Find the H3O of a solution whose pH4.0
• H3O 10-4 M .0001M
• Find the H3O of a solution whose pH3.54
• H3O 10-3.54 2.88 x 10-4 M
• Use the same procedure to find OH- from pOH

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Examples 5 and 6

• Example 5
• Find the H3O of a solution whose pH2.6
• H3O 10-2.6 2.51 x 10-3 M
• Example 6
• Find the OH- of a solution whose pH 5.8
• pOH14-5.8 8.2
• OH- 10-8.2 6.3 x 10-9 M

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OH- or H3O calculations

• Ex. 1 Find the OH- of a .25M HCl solution
• H3O OH- 1 x10-14
• .25 OH- 1 x10-14
• 1 x10-14/.25 4 x 10-14 M

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pH calculations Weak Acids and Weak Bases

• Weak acids such as acetic acid do not separate
  into ions 100 as do strong acids like HCl.
• The pH of a 1.0 M HCl solution is 0.00
• The pH of a 1.0 M CH3COOH solution is 2.37

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Ka

• The extent of ionization is measured by an
  equilibrium constant (Ka)
• If you use HA as the formula for a generic weak
  acid HA H2O ? ? H3O A-
• The Ka H3O A-/HA
• The represents molarity
• For CH3COOH (acetic acid)
• Ka 1.8 x 10-5

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R I C E table and Ka

• The method used to calculate pH from Ka is called
  rice for
• Reaction
• Initial conditions
• Change
• equilibrium

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Find the pH of a .50M CH3COOH solution

• R CH3COOH H2O ? ? H3O CH3COO-
• I .50 M 0.0M
  0.0M
• C -x x
  x
• E .50-x x
  x
• Then set up the Ka equation
• 1.8 x 10-5 H3O CH3COO-/CH3COOH
• 1.8 x 10-5 x2/(.50-x)

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Approximation method

• We will always decide that the x in the
  denominator of 1.8 x 10-5 x2/(.50-x) is too
  small to make a significant difference in the
  calculation so the equation simplifies from a
  quadratic equation to
• 1.8 x 10-5 x2/.5
• X (9 x 10-6)1/2
• x .003M
• pH -log (.003) 2.52

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Weak base calculations use Kb

• The method of solving a Kb question is the same
  as a Ka question
• Example Find the pH of a .450M NH3 solution if
  the Kb 1.8 x 10-5
• R NH3 H2O ?? NH4 OH-
• I .450M 0 0
• C -x x x
• E .45-x x x
• 1.8 x 10-5 NH4 OH- / NH3
• 1.8 x 10-5 x2/(.45-x) ? ? 1.8 x 10-5 x2/.45
• X 2.85 x 10-3 M
• pOH -log(2.85 x 10-3 ) 2.55
• pH 14 pOH 14- 2.55 11.45