Titration Calc: 0.01M NaOH & 10mL Ethanoic Acid pH

AI Thread Summary
The discussion focuses on a titration involving 0.01 M NaOH and 10 mL of ethanoic acid, with participants calculating the pH at the equivalence point. After determining the pH for the first titration, the conversation shifts to a second titration of 0.01 M NaOH with 10 mL of 0.200 M HOBr, a weak acid with a pKa of 8.69. Participants are tasked with calculating both the initial pH and the pH at the equivalence point for this second titration. A link to a pH calculation resource is provided for assistance. The thread emphasizes the importance of understanding pH changes during acid-base titrations.
Paulo2014
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A titration in which 0.01 mol L NaOH is added to 10ml of ethanoic acid is carried out.
find the pH at the equvilence point
 
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Lol thanks man.

The next part of the question says: A second titration is carried out in which 0.01 mol L NaOH is titrated with 10ml of 0.200 mol L HOBr, a weak acid with a pKa of 8.69.
Calculate the initial pH and the pH at the eqivilence point.
 
Thread 'Confusion regarding a chemical kinetics problem'

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...
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