What Is the pH at Equivalence for a Weak Acid-Strong Base Titration?

AI Thread Summary
In a titration of 25 mL of 0.1 M acetic acid (HOAc) with 0.15 M sodium hydroxide (NaOH), the pH at equivalence will be greater than 7 due to the weak acid-strong base reaction. To find the exact pH at equivalence, one must first calculate the moles of NaOH required, using stoichiometric relationships from the balanced chemical equation. The resulting solution at equivalence will contain sodium acetate, the conjugate base of acetic acid. The pH can then be determined by using the Kb of sodium acetate. This approach allows for an accurate calculation of the pH at the equivalence point.
monae
Messages
2
Reaction score
0
25 mL of 0.1 M HOAc are titrated with 0.15 M NaOH. What is the pH at
equivalence?

I know that the pH at start will be greater than 1.0 and the pH at equivalence will be greater than 7 since its a titration of a weak acid and a strong base. But what I don't know is whether or not there is a way to solve this problem for the exact pH since the volume of NaOH is not given.
 
Physics news on Phys.org
can't you figure out the volume of NaOH, though?

write and balance the chemical eqn, and then use stoichiometric relationships to find the moles of NaOH. Divide this by the molarity (0.15 M) to get milliliters of NaOH.
 
monae said:
25 mL of 0.1 M HOAc are titrated with 0.15 M NaOH. What is the pH at
equivalence?

I know that the pH at start will be greater than 1.0 and the pH at equivalence will be greater than 7 since its a titration of a weak acid and a strong base. But what I don't know is whether or not there is a way to solve this problem for the exact pH since the volume of NaOH is not given.

The pH at equivalence point is the same as solution containing the equivalent concentration of sodium OAc. You can use the Kb of this conjugate base of the weak acid.
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...

Similar threads

Replies
2
Views
1K
Replies
4
Views
1K
Replies
7
Views
7K
Replies
3
Views
2K
Back
Top