Question

In the titration of 25.00 ml of 0.10 M HF with 0.10 M NaOHn given ka = 3.5 x 10^-4 find

initial PH, PH at the equivalence point and Ph at half equivalence point

Answer 1

Grineni volume of HF - 25.00ml [HF) = 0.10 M kg-3.58164 [NaOH) = 6.10 Initially Finding pH: Hran Mt F- (99) (99) 0 (partial dissociation) 1 C tu th ofn (neaal} u 22 pe. [+] - ka= [W][Fy [HFI 35 x 10 - (+ ) (4) olan = n² = 3.5810 Хо: 1 = 1358 104 n=0.59X10+? M. 0.592X 10? 5.92 4103M pH = -log('ornyer dog [5.92 x 10"}] - 2. 23 Initial pH is 2.23 Ang pH at Half equivalent point = ? To find pH of at half equivalent point, we need to add exactly half the moles of beese to neutralize exactly half of the acid. HF + NaOH Na F + H₂O Initial solution of HF contenus not = 01) X.025 =0.0025 moles HF At half equivalent point solution meyt contain £8.0025= 1,25X103 moles, of HF volume of Nooth needed for 1,258103 moly 1.25 7 10 .] >> -3 - Liten - 12.5 mL

Total volume at half equivalent point - 2500L + 12.5 = 37.5 mL = 37.5x10 Liten. Concentration of HF at half equivalent point 1.25 X1003 37.5X10-3 - 0.033 HE -> H+ + F- o 4 0,033 tu +2 с th fr 0.033.n ka = (M) (2) 3.58104 0.033-2 (nac 0.033 ) = n² = 0.03383.58159. n =:34x102 ie. [HA] = 34x10m2 pH = -log[ht] = -log [-34x102] pH = 2.47 Ang at half equivalent point. ? =0.0025.025 Lite pH at equivalent point = at equivalent point male of acid = mole of base male of acid = 1/8.025=.0025 = mole of base valu HE & NaOH Na F + H2O volume of Naoh 0.0025 00025 required 0.0025 -0.002S -0.0025 ENAFE 0.0025 .625 4.025 = 25ML =.05M NOF Егие о НЕ 4 он os [F- uson and Kw = kako & k b = 3.58104 Total volume volymela cid + bases 100 oty Nattf to = [HF] [or 0 ܟܙ fn th n + tn 1044 22.8681511