What is the Ka and pKa for a weak monoprotic acid with given titration data?

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To calculate the Ka and pKa for a weak monoprotic acid using titration data, the key information includes the volume of NaOH used (55.48 mL) to titrate 25 mL of the acid and the pH (4.96) after adding 31.64 mL of NaOH. The relevant equations are pKa = pH + log(acid/conjugate base) and Ka = products/reactants. The concentration of NaOH is crucial for accurate calculations but was not provided in the problem statement. The original poster initially struggled with the approach but later found a solution. Understanding the relationship between the acid and its conjugate base is essential for determining the pKa and Ka values.
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Homework Statement


Kk so the problem is that we have 55.48 ml of NaOH was required to titrate 25 mL of a weak monoprotic acid. After 31.64 mL of NaOH had been added, the pH is 4.96 for the solution. Calculate the Ka and pKa for this weak acid.

Homework Equations


pKa= pH + log(acid/conjugate base) maybe?
ka=products/reactants

The Attempt at a Solution


I wanted to make the NaOh and acid a ratio and then since we know what pH is, just use the first equation but I don't think that'll be correct. Other than that I'm not sure how to approach this. Help will be aprreciated thanks
 
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Not enough info. What is the concentration of NaOH?
 
the concentration isn't given, I copied the exact problem.
 
nevermind, i found out how to solve it
 
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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