Question 43

[sunnyb321033]

Hi Can I please be offered an explanation for how you get the coefficient of 5 in front of Log when calculating for the pH.
Thank You

Thus, pH = pKa + log([A-]/[HA]) = 5 log(1) = 5.

how did u get the 5 in 5 log(1)? ??

[admin]

The calculation does have 2 steps not clearly written. Let's begin with what is mentioned in the passage:

Now, half as much NaOH as acetic acid is added; hence, in the end, [acetate] = [acetic acid].

Thus, pH = pKa + log([A-]/[HA]) = -log (1.8 x 10^-5) + log (1) = 5 log(10) = 5 approximately.

Reminder of your rules of logarithms: -log (1.8 x 10^-5) = -log 1.8 -log 10^-5 = -log 1.8 + 5log 10 = - log 1.8 + 5

[msamadia]

why does acetate=acetic acid from the knowledge that 0.005 mol of NaOH titrate acetic acid?

I don't understand.

[msamadia]

ohh is it b/c it is at the flat part of the graph not the long part so acid=base equal amounts?

[kpbakshi5858]

I dont get it. Wouldn't the pH be basic? Here's my reasoning:

1.8x10^-5=x^2/.01
[H+]=x= approx 4*10^-4
[OH-]=(.05L)(.1M)=5x10^-3

The H+ will react with the OH- until one of them is completely gone, and because H+ is the limiting reagent, there will be excess OH- left in the solution:
[OH-]=(5x10^-3) - (4*10^-4)= 4.6x10^-3

So then I do pOH=-log(4.6x10^-3)= 2.3
pH=14-pOH= 14-2.3=11.7

Is my reasoning wrong?

[dcherian8173]

You can think of the strong base as "driving" the formation of more acid. The strong base forms [OH-] and combines with the [H+] from our weak acid, producing H2O and thus removing the [H+] product from solution. As a result, the weak acid will produce additional [H+] to compensate, until all the [OH-] from our strong base is used up. At this point, 50% of our original acid has been disassociated into its anion.

[admin]

To solve for this one needs

1.To identify the major species in the solution which are acetic acid (a weak acid), sodium ion (a spectator ion), acetate ion (conjugate acid) and water (amphoteric).

2.To identify the equilibrium reactions involved which were: CH2COOH < --- > H+ + CH3COO-

3.To determine the equilibrium concentration as illustrated below






4. To determine the equilibrium constant expression which is Ka= [(conj base) (conj acid)/(acid)].

So the equation will be 1.8 x 10-5= [(0.01+x)(x)/(0.01-x)].

Assuming that x << 0.100 M because Ka x 1000 < initial concentration of acetate and acetic, then the equation will be just 1.8 x 10-5 = [(0.01)x/(0.01) where x= [H+].

Then pH will be pKa.

[H+]= Ka= 1.8 x 10-5

pH= 4.74

[admin]

This is a weak acid strong base titration and so you are essentially making a buffer solution at the midpoint of a titration curve for weak acids and strong bases. So for this titration, there is a buffer range that occurs before the equivalence point is reached due to the acid and conjugate base formation.

By using the reaction stoichiometry to compute the amounts of each buffer component and then using the Henderson-Hasselbalch equation we see the equivalence effect. Moreover, halfway to the equivalence point, the buffer components are exactly equal and the stoichiometry indicates that scenario. In addition, also note pH = pKa at that point so one may also determine the identity of the acid being titrated.

[admin]

Looks fine… However you should tackle this question using the IRF (or ICE) method in which you start with an initial concentration or moles of acid and base and as you add the base you will allow the amount added to react (usually) the amount would be a variable x and you set up the IRF or ICE table…. Then you need to divide the total amounts of moles of H+ and OH- reacted by the total volume added together. So it looks as though you did tackle it accordingly. Please let me know if you need the complete calculation by IRF…

[ecl3c7ic]

For acetic acid, A- and HA are, respectively, acetate and acetic acid; if it is true that [acetate] = [acetic acid], why is log([A-]/[HA]) not log(1)? It looks like the concentrations of A- and HA are equal, so I don't quite understand how [A-]/[HA] = 10.

[jellywing_2058]

As mentioned in the previous posts, some steps are skipped. (The value 10 does not represent [A-]/[HA])