1) The titration of a 25.0 mL sample of a base requires 23.6 mL of 0.100 M HCl.
ID: 625940 • Letter: 1
Question
1) The titration of a 25.0 mL sample of a base requires 23.6 mL of 0.100 M HCl. What is the concentration of the base assuming that it is mono-protic? 2) The titration of a 50.0 mL sample of di-protic acid requires 23.6 mL of 0.100 M NaOH. What is the concentration of the acid? 3) In the titration of a 25 mL sample of a 0.110 M solution of formic acid with 0.110 M strong base, what is the pH initially? 4) In the titration of a 25 mL sample of a 0.110 M solution of formic acid with 0.110 M strong base, what is the pH at the half-equivalence point? 5) In the titration of a 25 mL sample of a 0.110 M solution of formic acid with 0.110 M strong base, what is the pH at the equivalence point?Explanation / Answer
a) N1*V1 = N2 * V2 Since acid = HCl and base is mono - protic, N1 = M1 and N2 = M2 25 * M = 23.6 * 0.1 Concentration of Base = 0.0944 M b) N1*V1 = N2 * V2 Since acid us di-protic, N1 = 2 * M1 and base is mono - protic,, N2 = M2 2*50 * M = 23.6*0.1 Concentration of = 0.0236 M c) Initial pH = - log( 0.110) = 0.9586 d) pH = pKa + log [base]/[acid] where "base" and "acid" are the weak acid and its conjugate base. At half equivalence point, the concentrations of the weak acid and its conjugate base are equal. So that ratio is 1. The log of 1 is zero, so, the pH = pKa. pKa of Formic acid = 3.74 ( as per the data given in many chemistry books) e) At the equivalence point, you no longer have a solution of Formic acid, but you have a solution of the salt formed. Since your acid and base had the same concentrations, when you reach the equivalence point, you will have 25 mL of a solution of sodium acetate at a concentration of 0.055 M. Kb for this reaction can be calculated from KaX Kb = Kw to be 5.56 X 10-10. By that equation, [HAc] = [OH-], so Kb = x^2/0.055 You can solve this to show that [OH-] = 5.5 X 10-6 M. pOH, then = 5.2596 and pH = 14 - pOH = 8.74 Do rate the answers bro! Cheers :)
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