WCLN Ka to pH and Percent Ionization Chemistry

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Goes through the procedure of setting up and using an ICE table to find the pH of a weak acid given its concentration and Ka, and shows how the Percent Ionization (also called Percent Dissociation) is calculated. • http://www.BCLearningNetwork.com. • 0:00here will show you how to calculate the • 0:06pH and the percent ionization of a weak • 0:09acid with a given concentration and an • 0:1101 k value for the a part of this • 0:15question we are asked to find a ph of a • 0:17point 2 5 molar solution of methanoic • 0:20acid ch3cooh the be part of this • 0:24question asked to find the percent • 0:26ionization of the acid will start with • 0:30the a part we're asked to find the pH of • 0:32a point 2 5 molar solution of methanoic • 0:35acid • 0:36whenever asked to do a calculation with • 0:39an acid the first thing we always have • 0:41to do is identify the acid as strong or • 0:44weak we see by its position on the acid • 0:48table that ethanoic acid is a weak acid • 0:52when we're doing calculations involving • 0:55weak acids we must use a nice table • 0:58we set up a nice table like this • 1:01we start by writing the equilibrium • 1:03equation for ionization of methanoic • 1:06acid here • 1:07next we draw borders then so that the • 1:10columns line up nicely with the • 1:12substances in the equation • 1:14water is a liquid in equilibrium • 1:16equation so we can ignore the column • 1:19below water will color it blue here • 1:22we'll start out with the initial • 1:24concentration roll the initial • 1:26concentration of the ethanoic acid is • 1:29point 2 5 molar will add point 25 to • 1:32this cell • 1:33no h3o plus or ch3cooh minus was added • 1:37so we can consider their concentrations • 1:39to be zero before ionization • 1:42now we'll look at the changes in • 1:44concentration as ionization occurs • 1:47because there are no products initially • 1:49the equilibrium will shift to the right • 1:51during the ionization • 1:54as a shift to the right occurs the • 1:56concentration of hydronium and evaluate • 1:58ions will both increase will write plus • 2:00signs here and the concentration of • 2:03ch3cooh will decrease so well write a • 2:06minus sign here • 2:08we're not given any equilibrium • 2:09concentrations we don't know how much • 2:12these will increase so we'll write plus • 2:15x4 both of them as they both have a • 2:17coefficient of 1 in equilibrium equation • 2:21because the coefficient on ch3cooh is • 2:24also one we can state that it will go • 2:26down by X alright minus X here • 2:30now we'll look at the concentrations of • 2:32everything at equilibrium • 2:36the hydronium and evaluate ions will • 2:38both be 0 • 2:40+ x which is equal to x • 2:44the concentration of ch3cooh started out • 2:47as point 25 and went down by X so its • 2:50equilibrium concentration will be point • 2:5225 minus X the question now is how do we • 2:56find X we start solving for x by writing • 3:00AKA expression for ethanoic acid using • 3:03the equilibrium equation we see that the • 3:05KA expression is equal to the • 3:07concentration of hydronium times the • 3:09concentration of a found a way to over • 3:11the concentration of methanoic acid • 3:14the equilibrium concentrations of • 3:16hydronium and found a weight are both • 3:18equal to X so further product in AKA • 3:21expression we can substitute x times x • 3:24or x squared the concentration of • 3:27methanoic acid at equilibrium is point • 3:29two five minus X so we'll substitute • 3:32point 25 minus xn for the concentration • 3:35of CH 3 c.o wait • 3:38the degree of ionization for ethanoic • 3:40acid is very low so we make the • 3:43assumption that X is insignificant • 3:44compared to point two five • 3:47this can be written as point 25 minus X • 3:50is almost equal 2.25 • 3:53we can check this assumption later when • 3:55we determine the percent ionization in • 3:57general the assumption is valid if the • 3:59percent ionization is five percent or • 4:02less • 4:03using this assumption will help us avoid • 4:05having to use a quadratic equation and • 4:08chemistry 12 this assumption is • 4:10generally used here when you use it you • 4:13must always stated • 4:16so taking out the X on the bottom we can • 4:19state that k is approximately equal to x • 4:22squared over point 25 • 4:25rearranging this equation to solve for x • 4:27squared gives us x squared is equal to • 4:30point two five times the KA taking the • 4:33square root of both sides gives us x is • 4:36equal to the square root 2.2 five times • 4:39the KA • 4:40at this point we remind ourselves that x • 4:44is equal to the equilibrium • 4:45concentration of hydronium looking on • 4:48the asset table we see that the KA for • 4:51ethanoic acid is 1.8 times 10 to the • 4:53negative fifth so we substitute that in • 4:56for the KA in equation point 25 times • 5:001.8 times 10 to the negative fifth is • 5:03equal to 4.5 times 10 to the negative • 5:05six so we'll substitute that in here • 5:08under the square root sign • 5:10the square root of 4.5 times 10 to the

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