Question

1 of 20 Consider a buffer solution that is made out of hydrofluoric acid (HF: Ka = 3.5.10-) and sodium fluoride (NaF). The buffer rage of this buffer solution is: 2.5. 10-4 to 4.5.10 2.46 to 4.46 2.96 to 3.96 3.46 to 5.46

Answer 1

. bo Bage) A njugate al Soketion : Given Data!- Ka of HF: 3.5X1059 from Handerson- Hasselbalch equations we know that the pH. Base = conjugate base pH = pkät log Por ng Tacid? ? Alida conjugate and At half equivalence point the concentration of Base equal to the concentration of and le. [Base] = [Aud] so, the ratio of Borse & Acid is 1 and ph pH = pka + log (1) pH = pka - cij Case-I : Base is added 10 times greater than the aud so, T Bose ] = 10 Tarid] = 1 pH = pklat 1 - liis - pH = pka + log 10 Case -II - Suppose aid is added 10 times greater than Bose then pH = pka tlog 17 pH = pka - 1 - (11) [Base] =1 {Aud] =10 cs Scanned with CamScanner

- ciij Acil we conclude that from equation - pH= pka 21 e pH= pka pka= -log (ka) :-109(3.5X10-4) = 3.46 pH 3-46 + 1 4.46 pH = 3.46- 1 = 46 So, The pH range of this buffer is 2.96 to 4.46. option 'Ind is correct. Scanned with CamScanner