Provided is the question and answer. Please explain in full detail how the answe

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Provided is the question and answer. Please explain in full detail how the answer was obtained for study purposes.
Question 6 For the dissociation reaction of a weak acid in water, HA(aq) H20 (l) at H30 (aq) A (aq) the equilibrium constant is the acid-dissociation constant, Ka, and takes the form (H30 Weak bases accept a proton from water to give the conjugate acid and OH ions: B(aq) H2O() BH (aq) OH (aq) The equilibrium constant Kb is called the base-dissociation constant and can be found by the formula When solving equilibrium-based expression, it is often helpful to keep track of changing concentrations is through what is often called an I.C.E table, where I. stands for Initial Concentration, C, stands for Change, and E. stands for Equilibrium Concentration. To create such a table, write the reaction across the top creating the columns, and the rows l.C.E on the left-hand side, A B AB Initial (M) Change (M) Equilibrium (M) Part A Aspirin (acetylsalicylic acid, C9HsO4) is a weak monoprotic acid. To determine its acid- dissociation constant, a student dissolved 2.00 g of aspirin in 0.600 L of water and measured the pH. What was the Ka value calculated by the student if the pH of the solution was 2.61? Express your answer numerically using two significant figures. 38.10 Submit Hints My Answers Give Up Review Part Correct Part B A 0.100 M solution of ethylamine (C2HsNH2) has a pH of 11.87. Calculate the Kb for ethylamine Express your answer numerically using two significant figures. Submit Hints My Answers Give Up Review Part Correct

Explanation / Answer

let acetylsalicyclic acid be HA.

Concentration of HA given = moles/volume

Moles= mass of salicyclic acid/molar mass. Molar mass of C9H8O4= 9*12+8+64= 180

Concentration of HA = 2/180/0.6=0.01852 M

HA +H2O ------->A- +H3O+

given pH= 2.61, -log[H3O+] = 2.61, [H3O+] = 10(-2.61)= 0.002455 =[A-]

At Equilibrium [HA] = 0.01852-0.002455= 0.016065=1.6065*10-2

Ka= [H3O+] [A-]/[HA] = 0.002455*0.002455/0.016065= 0.000375

2. B( ethyl amine)+ H2O ------>BH+ +OH-

pH= 11.87, pOH= 14-11.87= 2.13, [OH-] = 10(-2.13)= 0.007413, =[BH+]

at Equilibrium [B]= 0.1-0.007413= 0.092587

Kb= [BH+] [OH-]/[B] = 0.007413*0.007413/0.092587= 0.000594=5.94*10-4

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