TBR, Gchem, Acids and Bases, Passage 5 # 34

[Sammy1024]

34. After 0.10 moles of an unknown weak acid HA are dissolved into 100 mL of H2O, the solution's pH is 3.7. What is the concentration of the conjugate base (A-)?

A. [A-] >2x 10^-4M
B. [A-l = 2 x 10^-4M
C. [A-] < 2x 10^4M
D. The Ka of the weak acid must be known to determine the value of [A-].

34. Choice B is correct. This question appears to be difficult at first glance. However, if you realize that when an acid dissociates, equal parts of hydronium and conjugate based are formed, then you know that [H30+] = [A-]. All that is required is converting from pH to [H30+], using [H30+] = 10^-PH. The pH of the solution is 3.7, so the [H30+] is 10^-7=1003x10"4. Because log2=0.3, it is true that 10^.3=2. This means that 10^.3 x 10^4 = 2x10^4. The [H30+] is 2x10^4M, and [H30+] = [A-], so [A-] = 2x10^4. The correct answer is choice B.

Can someone explain this answer to me? Thank you!

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[dudewheresmymd]

this is just algebra.

First, understand that HA <--> [H+][A-]

therefore concentration at equilibrium [H+]=[A-] (each are "x" in the ice box)

so if you can solve for [H+] you can get [A-], well they give you the pH so 3.7 =-log (H+)

10^(-3.7) =[H+]

Now it's just TBR math time, (they love these facts log(2)=.3 log(3)=.48)]

so they rearrange 10^(-3.7) = 10^(-4) * 10^(.3) [remember add exponents when you multiply] and therefore since we know log (2)=.3 (10^.3 =2 in other words) it's just 2 * 10^-4.

 

[Sammy1024]

Thank you so much!

 

[dudewheresmymd]

Thank you so much!

you got it

 

[drechie]

such a crappy move TBR