Quiz 1 #5

[Celeste 1I]

Can anyone tell me the correct answer for this question. I got it wrong initially and want to know if I've done it correctly this time. Thanks!

[Jim Brown 14B Lec1]

What day is your discussion, the answers are different depending on the day.

[Christopher Tran 1J]

If your question had a pKa value of 4.17 for 0.040M sodium ascorbate, the pH of the solution is 8.39.

[Celeste 1I]

oops didn't know there were different forms. The question asks for the pH of a .030M solution of sodium sorbate(C6H7O2), given that the pKa of sorbic acid(C6H8O2) is 4.74. The reaction is: C6H7O2+H2O<=>C6H8O2+OH

[bryan de leon disc1k]

Ka was equal to 6.76x10^-5
kb=1.48x10^-10

[bryan de leon disc1k]

Ka was equal to 6.76x10^-5
kb=1.48x10^-10

[Sarah Bui 2L]

I think each discussion had similar questions, so I am not sure if we have the exact same question, but I can go over the steps. What is given is the reaction and since one of the products is OH-, you know that sodium ascorbate (or whichever it says on your test) is a base. Because what they give is pka, we want to change this into pkb (since we are working with base). To find this we subtract 14-pka. You can then get the Kb value from pkb by doing Kb =10^-pkb. Then set up your ice table, and input your values (intial concentratio of your base and zero concentration for your products). From then just solve for x, you can also do the approximation since your Kb value is less than 10^-3, but make sure to use the <5% rule to check. Your x value will tell you the OH concentration, plug into -log(OH). BUT, since they ask for pH, then pH=14-pOH.

[Christopher Tran 1J]

oops didn't know there were different forms. The question asks for the pH of a .030M solution of sodium sorbate(C6H7O2), given that the pKa of sorbic acid(C6H8O2) is 4.74. The reaction is: C6H7O2+H2O<=>C6H8O2+OH

Ka = 1.82*10^-5
Kb = 5.50*10^-10

x = [OH-] = 4.06*10^-6
pOH = 5.39

Therefore, pH = 14-5.39 = 8.61