Determine pH and POH From the Given Equation

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• Understanding the Concepts of pH and pOH, and the Relationships Surrounding Concentration, Moles, Molarity, and Number of Molecules

  pH = -log[H+] pOH = -log[OH-] pH + pOH = 14 From the question, we know that we have 250 mL of a pH 3.0 solution.

• Finding the Concentration of Fe2+ in equilibrium with Fe(OH)3

  Fe(OH)3 = 4 x 10-38 4 x 10-38 = [Fe3+] * [ OH-]3 Given that , pH = 6.00 pH + pOH = 14.0 pOH = 14.0 - pH = 14.0 -6.00 = 8.00 pOH = -log[OH-] [OH-] = antilog[-pOH] = antilog [-8.00] = 1 * 10-8 moles

• Determine the pH of a 1.0 L of the buffer that is 0.50 M in NH3 and 0.20 M in NH4Cl after the addition of 0.010 mol of NaOH.

  492957 Determining the pH of the Buffer Determine the pH of a 1.0 L of the buffer that is 0.50 M in NH3 and 0.20 M in NH4Cl after the addition of 0.010 mol of NaOH. Kb = 1.8 x 10^(-5) for NH3. The calculation should be in two parts.

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  Define pH and hence calculate the pH of a solution of a) 0.01 M hydrochloric acid b) 0.1 M sodium hydroxide c) 0.2 M sulphuric acid d) 0.1 M butanoic acid, given that Ka = 1.51 x 10-5 Explain clearly the difference between a strong acid and a

• PH Calculation Basic Principles

  + (-log10[OH-] ---------- (ii) Note: -log10Kw = PKw ------------ (iii) -log10([H+] = PH ----------- (iv) -log10[OH-] = POH ------- (v) Substituting equations (i), (ii) and (iii) in equation (ii), we have PKw = PH + POH Solution

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  the Kb is so small, you can use the [OH-] to find pOH and pH.

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• pH, pOH

  pH = -log[H] = 14 - pOH pOH = -log[OH] = 14 - pH Just need to reverse when the other is given.

• pH of Sodium Acetate at Room Temperature

  Thus from the above expressions for equilibrium concentration for OH-, we know that the concentration of OH- is 2.36 x 10 -5 M and the pOH is pOH = - log ([OH-]) = - log (2.36 x 10 -5 M) = 4.63 and thus the pH = 14 - 4.63 = 9.37.