Question

Strong base is dissolved in 765 mL of 0.200 M weak acid (Ka = 3.76 *105) to make a buffer with a pH of 4.03. Assume that the volume remains constant when the base is added. HA(aq) + OH(aq) → H20(1)+ A- (aq) Calculate the pKa value of the acid and determine the number of moles of acid initially present. Number Number PK, = 10 mol HA When the reaction is complete, what is the concentration ratio of conjugate base to acid? Number TA How many moles of strong base were initially added? Number mol OH

Answer 1

tot part 0.2(M) weak acid solution means that - 1000 ml volume solution contains 0.2 moles of Weak acid. 1 " 19 12 11 0. 2 2 in 1000 76 5: 1 1 » 1 0.2776.5 1000 . . . = 0.153 moles of . . weak acid. In 765 ml of 0.2(M) weak acid, the number of moles of weak acid initially present = 0.153 moles: ka = 3.76x105 ... from the definition of Pka = -logiska ! pkas - log (3.76X105) . pka, = 4:425 : and part : An amount of strong base is added in 765 ml of 0.2 (M) weak acid. It becomes acidie buffer. Henderson-Hasselbalch equation for acidic buffer, pH = pkai t log [salt] & Tacid The rean, HA (99) + Ort (ag) → H₂0 (1) + A (ar). Here is the conjugate base of weak acid HA.. - -

E. At the same time , A. is the ion coming from the salt formed in this titration reaction (Weakaid-strong base), 80, [salt] = [A-] and [HA] = It is the concentration of unreacted HA acido TAT : Now, PH₂ pka + log LA . (Here pH= 4:03; pha= 4.425) on, 4.03 = 4.425.+ long and or, long China = 4.03 -4.425 = – 0.395 Or, [-] = 100.395 H47. [A-] = 0.403 [HA] = 0.403 . after the reaction, [A-] 2.0.403 . the THAJ -

3rd part: 0.403 concentration no, of moles (n) THAI volume or, n I Here volume of the solution = 0.403. remains constant during Landego addition of the base, o ANA-ano, of moles of A ion(from) and nha 2 no, of moles of the unreacted HA * present in the solution. ? MHA NHA or, ... NA TNHA MHA Now nA tla 17 0:403 | (MA + MHA)= 0.153 It is equal to the = 1.403. I of ha before strong initial no. of moles. base addition, a according to the chemical equations provided) : i mol of HA reacts with Imol of óh/strong base) giving 1 mol of A-(salt). .. NA moles of HA reacts with nA moles of õh ione giving nA moles salt (A-). (no. of moles of HA reacted with 5k) + (no. of moles of unreacted HA in solution) = initial .. no. of moles of HA

- :. - + HA = 0.153. na thHA 1402 NHA or, 0.153 the value of. = 11403 Ing-thHA = 0.153 NHA is put NHA = 0.153 LMHA= 0.109 Putting the value of NHA in eq: , we get. na :=0.403 . 1.403 NHA on naa 0.403 NHA or MAZ. 0.403.x 0.109 i Maa 0.0439 0.0439 moles of strong base were initially added that is why we got 0.0439 moles of salt (A).