Wednesday

November 26, 2014

November 26, 2014

Posted by **simi** on Monday, March 1, 2010 at 11:38pm.

calculate the pH before titration

calculate the volume of added base to reach the end point

calculate the pH halfway to the equivalence point

kb(But-)= 6.6x 10^-10 and ka(HBut)= 1.5x 10^-5

calculate the pH after adding 5ml of NaOH beyond equivalence point

- chem -
**DrBob222**, Monday, March 1, 2010 at 11:50pmHow much of this do you understand and where do you get stumped? I can help you through the rough spots if you identify them but I don't intend to do all of your work for you.

- chem -
**simi**, Monday, March 1, 2010 at 11:57pmI understand how to get the pH before titration and how to get the volume. From my understanding to get the pH before titration i just multiply the ka and the weak acid concentration then i get the square root. I then get the negative log of the square root and i got 5.8 as the pH. For the volume I multiplied the volume by .105, i got .002 which i divided by .125 to get .016 as the volume. I do not know hoe to do the rest.

- chem -
**DrBob222**, Tuesday, March 2, 2010 at 12:23amThe procedure for finding pH at the beginning is correct but you made a math error somewhere. It appears you may have used Kb and not Ka.

1.5 x 10^-5 = (H^+)^2/0.105 and solve for (H^+). The pH is between 2 and 3.

Volume to reach the equivalence point is

mL x M = mL x M for the equivalence point volume of titrant.

pH at the half way point. You can do it the hard way by taking half the volume x M and find moles NaOH, then use up that many moles of butyric acid, subtract from initial butyric acid to find amount remaining, then plug those moles into the Ka expression and solve for H^+ and pH. Or you can do it the easy way and say to yourself--At the half way point, I have exactly half of the acid left and I have formed an equal amount of the salt; therefore,

(H^+)(Bu^-)/(HBu) = Ka and rearrange to solve for (H^+).

(H^+) = Ka x (HBu)/((Bu^-) BUT since (HBu)=(Bu^-), then (H^+) = Ka and pH = pKa.

The pH 5 mL beyond the equivalence point:

M x L = moles NaOH in excess. Then moles/total volume at that point gives molarity OH^-. You get pOH and pH from that.

- chem -
**simi**, Tuesday, March 2, 2010 at 12:49amoh yes i forgot to do the square root i get 2.9.

so for the second one it would be 4.8

and the third one it would be 10.8

thank you so much for explaining it step by step i appreciate the help

- chem -
**simi**, Tuesday, March 2, 2010 at 1:06amwhat about the pH at equivalence point?

- chem -
**DrBob222**, Tuesday, March 2, 2010 at 2:44pmI didn't see that the pH at the equivalence point was in the question. I just looked again and it isn't. But the pH at the equivalence point is determined by the hydrolysis of the salt.

Bu^- + HOH ==> HBu + OH^-

Kb = (Kw/Ka) = (OH^-)(HBu)/(Bu^-)

I think Kb is listed in the problem so it will not be necessary to calculate it with Kw/Ka. So substitute x for OH and x for HBu. For the Bu^-, that is the molality of the sodium butyrate at the equivalence point. You get that by moles HBu at the beginning will form that many moles of NaBu at the equivalence point and then you divide by the total volume (volume HBu initially + volume of NaOH added to get to the equivalence point). All of this gives you the OH from which you get pOH and pH.

- chem -
**Anonymous**, Tuesday, March 2, 2010 at 3:27pmokay thank you so much

**Answer this Question**

**Related Questions**

Chemistry - A student weights 1.700 g of succinic acid and dissolves it in water...

Chemistry - This is for a titration, to measure the amount of acid that can be ...

Chemistry - In a experiment to determine the molecular weight and the Ka for ...

11th grade CHEMISTRY TITRATION LAB! - i have a tritation lab due tomorrow and i ...

Chemistry - Calculate the pH at the following points during the titration of 100...

Chemistry - I don't know how to reply to an answer. First of all thank you! For ...

Titrations (chem) - In the titration of a 0.101 M sodium hydroxide solution, 19....

college chemistry - Consider the titration of a 50.0 mL sample of a 0.100 M ...

Chemistry - A 25.0mL sample of a 0.100M solution of acetic acid is titrated with...

chemistry - 11.During an acid-base titration, 25 mL of NaOH 0.2 M were required ...