The Acid Dissociation Constant, K added. As In the beginning of the titration th
ID: 559581 • Letter: T
Question
The Acid Dissociation Constant, K added. As In the beginning of the titration the pH changes very little as base is ate base of the base is added the salt NaCaH,O2 is formed. CHO2 is the conjuga HCH,02. Since HGH,0 is a weak acid then GHO·must be a relative good proton acceptor. Therefore, we have made a solution that contains a weak acid and its conjugate base. This solution is known as a buffer solution, Buffers solutions resist changes in pH they are diluted with H,O. We can understand how buffers work by considering the equilibrium present in the solution. The chemical reaction which describes the equilibrium present in a buffer solution can be written in generic terms as HX(aq) + H20(1) H,0.(aq) + X(aq) Here is HX is the generalized formula of a weak acid and X- is its conjugate base. The pH of a buffer solution can be calculated using the Henderson-Hasselbalch relationship pH=pK, +log(laesel) acid where pK, - -logK, and the terms [base] and [acid] refer to the concentrations of the conjugate acid and base. The pK, of the acid can be determined if the log term were to disappear, i.e. become zero nt tocten have to be for the log term to go to zero? Right...I's! Tbase] [acid] What would the quotient when the quotient equals 1 . At this point pH = pK, When does the quotient (basel become 12 Hint it's not at the equivalence point. To help [acid] you answer this remember that the base represents the conjugate base not the NaOH. The conjugate base concentration is the same as the concentration of the salt NaC2H302. So when does the concentration of the salt equal the concentration of the acid. To answer this imagine that you react 1 mole of HC H302 with 1 mole of NaOH. When does the concentration of NaC H 02 equal the concentration of NaOH? Using the graph and the information from the box above determine the pK, of the acid. Now calculate the KExplanation / Answer
The value in which
pH = pKa + log(base/acid)
is of interet:
base = acid = 1
is given at the 50% of the titration since 50% is converted to base, and 100-50 = 50 % is left
therefore, in this point
pH = pKa + log(1)
pH = pKa
then
from your graph
Vequivalence = 23 mL approx
half equiv point -- >23/2 = 11.5 mL
then
pH at this point
pH = pKa
pH = 4.5
pKa = pH = 4.5 approx
Ka = 10^-pKa = 10^-4.5 = 3.1622*10^-5
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