Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution. b. a solution that is 0.16 M in NH3 and 0.22 M in NH4Cl

Video transcript

Hello in this problem we are asked to use the ice table method to determine the ph of a solution containing 0.4 to moller ethyl amine and 0.64 moller methyl ammonium bromide forgiven based association constant for ethyl amine. So start then let's write the equation that describes the dissociation of our salt. So we have that's all morning bromine and associates the form ethyl ammonia mines and bromide ions. We're told the concentration of ethylene bromide is .64 Mueller. For every mole of ethyl bromide that associates we get an equal amount of alimony mines. So the concentration will be the same in our next step. Let's generate an ice table And so we're told the equilibrium constant for ethyl amine. So that's right. The analysis reaction or ethyl amine. So ethyl amine is a weak base. It will then accept a proton mater reform school ammonia minds which is the conjugate acid of our weak base and will form hydroxide ions completing our ice table. We have initial change and equilibrium. So our initial concentration of ethyl amine is 0.4 to Mueller will ignore the water since it's a pure liquid concentration of the ethyl ammonium bromide is initially 0.6 for Mueller and initially we have no hydroxide are changes minus X plus X plus X equilibrium. And then we combine the initial and the change and then the next step let's use our equilibrium constant solve for x our equilibrium constant expression. Then we have the concentration of our products divided by the concentration of are reacting and water being a pure liquid does not appear in the equilibrium constant expression using our ice table. This works out to 0. plus X Times X. All over 0.42 -X. And this is equal to 5.6 Times : -4. So we're gonna check our simplifying assumption to see if we can then simplify this calculation to eliminate the minus X. And the plus X. So to check are simplifying assumption. Then we're going to take the concentration of ethel mean initially divided by r based association constant. This then works out to 0.42, Divided by 5.6 times 10 to the -4. This works out to 750 Which is greater than 500. So we can simplify our calculation by getting rid of the -X and the plus X. So this then work out 2.64 times x. over 0.42 Is equal to 5.6 times 10 to the -4. And then solve for X. So I multiplied by .42 and divide by .64. And so X. Then works out to 3.675 times 10 minus four from our ice table. Then we know that X is equal to the concentration of our hydroxide ions. And our last step then we're gonna solve for P. H. And so we begin with the p O. H. P O. H. S. Then the negative log of our hydroxide iron concentration. This works out to 3.435, And then the ph is equal to 14 -300. So our ph works out to 10.57, So the p four, our solution is 10.57, and this corresponds to answer b. Thanks for watching. Hope this helps.