Calculate the concentrations of all species present and the
pH i...

Question

Calculate the concentrations of all species present and the 
pH in 0.10 M solutions of the following substances. See 
Appendix C for values of equilibrium constants. 
(b) Sodium acetate, Na1CH3CO22

Accepted Answer

Hello in this problem we are asked to find the ph of 0.125 molar sodium bentonite solution and the concentration of all species present in the solution. And to note that the K. Value for beauty in OIC acid is 1.48 times 10 minus five. Let's begin by determining whether our salt is acidic, basic or neutral. Looking at the formula for our salt, cat iron from our salt came from sodium hydroxide which is a strong base. So Dean martin then will result in the formation of a neutral solution. Our anna in came from weak acid and this is then our conjugate base of that weak acid. It will form a basic solution. Next let's make use of a nice table and are based association constant to find hydroxide ion concentration. So creating a stable We first need our equation. So we have are anNA. And from our assault which is a conjugate base of a weak acid. It will undergo hydraulic sis. We will form benton OIC acid and the dark side lines we have initial change and equilibrium Initially. Then the concentration of the Benton Oy is 0.125. We'll ignore the water since it's a pure liquid and we initially have none of the acid or hydroxide ions are changes minus X plus X plus X. We combine the initial and the change to get the equilibrium. Sorry equilibrium process expression then is equal to then our product concentrations over are reacting concentration. Water being a pure liquid does not appear in our equilibrium constant expression. And so we can find our based association constant. It's equal to then the iron product constant. Water provided by the acid dissociation constant product constant. Water is one times 10 to minus 14. And we were told that that's the dissociation constant for beauty. No gas is 1.48, 10 minus five because it tells us that our K. B value will work out to 6.75, 7 Times 10 to the -10. Writing our equilibrium constant expression in terms of the values from the ice table, we get X times X Over 0.125 -6. This is equal to 6.75, 7 times 10 to -10. I'm gonna check our simplifying assumption to see if we can get rid of the minus X. So we're gonna take our concentration of trying to get base provided by that for provided based association constant. We get 0.125 divided by 6.757, 10 to minus 10. This works out to 1.85 Times 10 to the eight, Which is much better than 500. So we can then simplify our calculation. So we end up then eliminating the -6. We get X squared. Then it's equal to 6.757 times 10 to minus 10 Times 0.125. When we move that to the other side. This thing gives us that X squared is equal to 8.44. 6 Times 10 to the -11. We'll take the square root of both sides. We get X. N. Is equal to the square root of 8.446 times 10 : -11. This works out to 9.190. I'm sending my six. And from the ice table this tells us then the concentration of hydroxide ions so we can find our P. O. H. This is equal to then the negative log of our hydroxide iron concentration. R. P. O. H. works out to 5.037. R P. H. C. N. Is equal to 14 minus P. O. H. This works out to 8.963. And from nice table, you know that our hydroxide ion concentration is equal to that of acid which is my point 10 : -6 Moller. And our concentration of The d. 1808 is equal to 0.125 moller. The study mind concentration will be the same is the concentration of future No eight. And our hydrogen mind concentration you can find go to attend to the negative ph This be tender negative 8.963 Which is equal to 1. 10 : -9 Moeller. So we have ph of our solution and then the concentration of the chemical species that will be in solution. This thing corresponds to answer D. Thanks for watching. Hope this helps

Calculate the concentrations of all species present and the pH in 0.10 M solutions of the following substances. See Appendix C for values of equilibrium constants. (b) Sodium acetate, Na1CH3CO22

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