WebExpert Answer. Transcribed image text: [1] A group of Aggie spudents prepared vibegar in the lab by using an initial concentration of Acetic acid of 0.25M. The dissociation of the acetic acid is shown below. If the Ka of the acetic acid was 4.8× 10−4, calculate the equilibrium concentration of [CH3COOH3][H3O+].[CH3COO∘]. pH and [OH2] of ... WebThe room temperature was given as 25°C. For the acetic acid solution, the Ka calculation for the 0.3M solution is calculated by using the equilibrium expression: Ka = [H+] [CH3COO-]/ [CH3COOH]. The initial concentrations of the reactants and products are given, and the change in concentration is calculated as the initial concentration minus ...
7.12: Relationship between Ka, Kb, pKa, and pKb
WebOct 23, 2024 · Hey Itzel, pK a is when you take the -log 10 of your K a.Therefore, taking the -log 10 (1.8*10-5), we get a value of approximately 4.74 for our pK a for acetic acid.. Hope … WebSolution. The Ka for oxalic acid is. K a = 1 E − 14 1.8 E − 10 = 5.6 E − 5. DIscussion. The calculation regarding Ka and Kb conversion is simple, but understanding what problems require this type of conversion is difficult. The concept is rather useful, and it further broadens the concept of acid and base. iron on flock vinyl
Worked example: Calculating the pH after a weak acid–strong …
WebThe pH of the buffer is 3.98. What is the ratio of the equilibrium concentration of sodium acetate to that of acetic acid if you assume that the Henderson-Hasselbalch equation is accurate? Question: A buffer is made with sodium acetate ( CH3COONa ) and acetic acid ( CH3COOH ); the Ka for acetic acid is 1.80×10−5 . The pH of the buffer is 3.98. http://clas.sa.ucsb.edu/staff/Resource%20Folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf WebApr 20, 2015 · Problem: A buffer was made by mixing 500 mL of 1 M (mol/L) acetic acid and 500 mL of 0.5 M calcium acetate. What are the resulting concentrations of acetic acid, Ca2+ and acetate and what is the resulting pH? Ka(acetic acid) = 1.75×10−5. Final volume = $500+500$ mL = 1L port owen camping