chem
posted by simi .
Consider the titration of 20 mL of .105M Butric Acid HBut, with .125M sodium Hydroxide?
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
How 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
I 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
The 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 yourselfAt 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
oh 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
what about the pH at equivalence point?

chem 
DrBob222
I 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
okay thank you so much
Respond to this Question
Similar Questions

Titration/Acidbase chem
a .4000 M solution of nitric acid is used to titrate 50.00 mL of .237 M barium hydroxide (Assume that volumes are additive). a) Write a balanced net ionic equation for the reaction that takes place during titration. b) what are the … 
Chemistry
In a experiment to determine the molecular weight and the Ka for ascorbic acid (vit. c.) a student dissolved 1.3713g of the monoprotic acid in water to make 50 mL of solution. The pH was monitored throughout the titration. The equivalence … 
chemistry
I am confused on how to do the last question. Calculate molarity of HCl from the volumes of acid and base at the equivalence point and the molarity of NaOH from the titration curve. (M of Acid)x(v of acid)= (m of base)x(v of added … 
chemistry
Consider the titration of 21.0 mL sample of 0.110 M HC2H3O2 with 0.130 M NaOH. what is the initial pH? 
Chemistry
In this case, the inflection point, and equivalence, occurs after 23.25mL of 0.40 M NaOH has been delivered. Moles of base at the equivalence point can be determined from the volume of base delivered to reach the equivalence point … 
Chemistry
A titration is performed by adding .600 M KOH to 40.0 mL of .800 M HCl. Calculate the pH before addition of any KOH. Calculate the pH after the addition of 5.0 mL of the base. Calculate the volume of base needed to reach the equivalence … 
chemistry
11.During an acidbase titration, 25 mL of NaOH 0.2 M were required to neutralize 20 mL of HCl. Calculate the pH of the solution for each of the following: 12.Before the titration. 13.After adding 24.9 mL of NaOH. 14.At the equivalence … 
Chemistry
A student weights 1.700 g of succinic acid and dissolves it in water in a 250.0 mL volumetric flask. A 25.00 mL sample of this solution is withdrawn and placed in a 125 mL Erlenmeyer flask, to which 5 drops of the acidbase indicator … 
Chemistry
1.) Watch the animation, and observe the titration process using a standard 0.100 M sodium hydroxide solution to titrate 50.0 mL of a 0.100 M hydrochloric acid solution. Identify which of the following statements regarding acidbase … 
chemistry
Consider the titration of 40.0 mL 0.250 M ethylamine, C2H5NH2, with 0.350 M HCl. Determine each of the following and sketch the titration curve. Kb of ethylamine = 5.6x 104 a. The volume of added acid required to reach the equivalence …