Determine [H3O+] and [OH-] in a solution with pH = 9.00. Question options: 1) [H

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The pH scale is a logrithmic scale. The p" factor" is defined as the log of the molar concentration of whatever follows the letter p and then multiplied by a negative So the pH = -log[H+] pCl = -log[Cl-] pKa = -log Ka pKw = -log Kw pAg = -log[Ag+] For strong acid molar concentrations equal to or less than 1, the pH value would have a value from 0-14. In a water solvent based solution: [H+] [OH-] = Kw = 1 X 10-14 If the Hydrogen ion concentration is 0.1 moles/liter Then the [OH-] could be found by the equation above: [OH-] = 1 X 10-14 / 1 X 10-1 = 1 X 10-13 The pOH = -log[OH-] = -log(1 X 10-13) = -(log 1 + log 10-13) = -(0 + -13) = -(-13) = 13 For a [H+] = 0.1 = 1 X 10-1 Then pH = -log 1 X 10-1 = -(0 + -1) = 1 Therefore the pH + pOH = 14 One can have a pH that is a negative value in for example strong acid solutions greater than 1 mole/liter. For a 10 M H+ pH =-log(1 X 101) = -(0 + 1) = -1 Most acidic and basis substances whose pH lie between the 0-14 range. Do not confuse the term acid strength with pH. The strength of an acid has to do with the percentage of the initial protons that are ionized. If a higher percentage of the original protons are ionized and therefore donated as hydrated protons (hydronium ions) then the acid will be stonger. Strong acids are Hydrochloric, Hydrobromic, Nitric, Sulfuric, and Perchloric acids. In each of these molecular acids the percentage of ionization is almost 100%. Determination of pH pH = -log[H3O+] The negative sign is to be multiplied by the log answer you get. Example: What is the pH of a solution whose [H3O+] = 1 X 10-4 M pH = - log[1 X 10-4] = - [ log 1 + log 10-4 ] Note: When you multiply numbers you always ADD their log forms log 1 is always zero log 10x = x so log 10-4 = -4 therefore: pH = - [ 0 + (-4) ] = - [-4 ] = +4 Here is a little harder Example: Calculate the pH of a solution that has a [H3O+] = 2.5 X 10-9 pH = - log [H3O+] = - [ log 2.5 + log 10-9 ] log 2.5 can be determined using a calculator having the log function key: Enter the number in this case 2.5 depress the log key Read the display which in this example would be .3979 log 10-9 = -9 so pH = - [ .3979 + -9 ] = - [ - 8.602 ] = + 8.602 Another example involving a base: Calculate the pH of a solution that has a [OH-] = 1 X 10-5 M Determine pOH pOH = - log [OH- ] = - [log 1 + log 10-5 ] = - [ 0 + -5 ] = - [-5] = +5 Determine the pH knowing that pH + pOH = 14 pH = 14 - pOH = 14 - 5 = 9 Determining the [H3O+] when given the pH Example: What would be the [H3O+] of a solution that has a pH = 5.4 [H3O+] = Antilog (-pH) = Antilog (-5.4) = 3.98 X 10-6 Note: Determining Antilogs is done in the following manner using a calculator: Enter the number which you wish to take the Antilog of in this case -5.4 (press the change sign key (+/-) to change the sign to negative) Depress the second function key sometimes called the inverse (inv) key Depress the log key Read the display Determining the [OH-] when given the pOH Example: Calculate the [OH-] of a solution that has a pOH = 8.2 [OH-] = Antilog (-pOH) = Antilog ( - 8.2 ) = 6.31 X 10-9 Determining the [H3O+] when given the [OH-] Example: Calculate the [H3O+] when the [OH-] = 3.2 X 10-3 Water ionizes only slightly giving the following equilibrium: H2O (l) + H2O(l) = H3O+(aq) + OH-(aq) According to the Law of Chemical Equilibrium: Kc = [H3O+] [OH-] / [H2O(l)]2 Note: concentration of pure water is a constant 55.5 Since the molar concentration of water is constant we multiply both sides of the above expression by [H2O]2 that will result in another constant: Kc [H2O]2 = Kw = [H3O+] [OH-] = 1.0 X 10-14 Kw = 1 X 10-14 = [H3O+] [OH-] Note: Kw is the ionization equilibrium constant of pure water which always has a value of 1 X 10-14 at 25 degrees Celsius. Kw = 1 X 10-14 = [H3O+] [ 3.2 X 10-3] [H3O+] = 1 X 10-14 / 3.2 X 10-3 = 3.125 X 10-12 Does it make a difference which acid or base is given in the problem? Yes, if the acid or base is a strong one, then the dissociation will be 100% For example, HCl is classified as a strong acid so: HCl + H2O ----> H3O+(aq) + Cl- (aq) If I start with a 0.1 M HCl, then I will have 0.1 M H3O+ because for every HCl that breaks apart then one H3O+ is formed and one Cl- is formed. Since strong acids (and bases) ionize 100% then all of the original concentration will be converted to H3O+ Typical strong acids: HCl, HBr, HI, HNO3, H2SO4, HClO4 Strong Bases: All the Hydroxide compounds of Group 1 and Group 2 metals LiOH, NaOH, KOH, etc and Be(OH)2, Ca(OH)2, Mg(OH)2, etc What would be the [OH-] of a .2 M NaOH solution? NaOH + H2O ----> Na+(aq) + OH- (aq) .2 M NaOH will produce .2 M OH- since the breakdown is 100% What would be the [OH-] of a .3 M Ca(OH)2 Ca(OH)2 + H2O -----> Ca+2 (aq) + 2 OH- (aq) .3 M Ca(OH)2 will produce .6M OH- because for every one Ca(OH)2 that breaks apart TWO OH- ions are produced (twice as much) (look at the equation and note the coefficients)

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