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pH calculations
pH calculations
pH calculations
pH calculations
pH calculations
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pH calculations

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PowerPoint to teach pH calculations for South African CAPS syllabus.

PowerPoint to teach pH calculations for South African CAPS syllabus.

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  • 1. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net pH Calculations K Warne
  • 2. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Acid strength & pH H3O+ concentration (mol/dm3) 1 0.01 0.001 Strong acids Weak acids 0.0001 0.000001 (1x10-4) (1x10-5)  Diluting a strong acid by a factor of ten only changes its pH by one unit. pH = - log10[H3O+] WEAK STRONG [H3O+] (mol/dm3) Log [H3O+] pH 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 B A S E SA C I D S W A T E R WEAKSTRONG NEUTRAL 1 0.1 0.01 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-12 10-13 10-14 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 -11 12 -13 -14 If acids produce H+ ions in solution then STRONG ACIDS must have a HIGH H+ ION CONCENTRATION. One mole in a liter (dm-3)is a high concentration. [H+] =10-pH
  • 3. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net pH Calculations – strong acids What is the pH of a solution with [H+] = 1.0 x 10-5 M ? pH = -log[H+] = -log 10-5 = -(-5.00) = 5.00 Show why the pH of pure water is 7. [H+]= 1 x 10-14 pH = -log[H+] = -log 10-7 = -(-7.00) = 7.00 What is the pH of 0.020 M HCl(aq)? Since HCl is a strong acid, it completely ionizes to give H+ and Cl-. HCl(aq)  H+(aq) + Cl-(aq) So: [H+] = 0.020 M pH = -log(0.020) = -(-1.70) = 1.70 Finding [H+] given pH What is the [H+] of a solution of pH 5.20? [H+] =10-pH =10-5.20 =6.3 x 10-6 M
  • 4. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Ionisation of water Water can ionise according to the following equation: H2O  H+ + OH- The [H+] and [OH-] are therefore equal. Since [H+] = 10-7 mol.dm-3 then [OH-] = 10-7 mol.dm-3 And [H+] x [OH-]= 1 x 10-14 This expression is called the ionic product of water (Kw) and it remains constant. (At a constant temp. @ 250 C) We can use this to calculate the pH of a base. If a base solution had a [OH-]= 10-3 Since [H+].[OH-]=10-14 then [H+] = 10-14/[OH-] [H+] = 10-14/ 10-3 = 10-11 M pH = -log [H+] = -log 10-11 = 11
  • 5. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Hi - This is a SAMPLE presentation only. My FULL presentations, which contain a lot more more slides and other resources, are freely available on my resource sharing website: www.warnescience.net (click on link or logo) Have a look and enjoy! WarneScience

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