Question

a) A 35.00 mL of 0.25 M HF is titrated with a 57.00 mL of standardize 0.1535 M solution of NaOH at 25°C. Calculate the pH at the equivalence point. Ka for HF = 6.76 x 10-4. (8.5 marks) b) What is the pH at 25°C of the solution obtained by dissolving an aspirin tablet in 0.500 L of water? The tablet contains 0.325 g of acetyl salicylic acid, HCgH704. The acid is monoprotic, and Ka equals 3.3 x 104 at 25°C. (7.5 marks) c) The following reactions are important environmental processes. Identify the conjugate acid-base pairs for the following reaction: H2PO4 (aq) + CO3(aq) HPO42- (aq) + HCO3 (ag) (2 marks) d) Predict whether aqueous solutions of the following are acidic, basic, or neutral, and write an equation for the reaction with water: i) Sodium benzoate, C6H5COONa ii) Ammonium chloride, NH4CI (4 marks) e) Decide if NaCIO4/HCIO4 can act as a buffer. (3 marks)

Answer 1

a) 35mYX 0.25 mol HF = 8.75 mmol #F y HF + Naot NaF + H₂O (1:1) My = M₂ V 2 arid Base 0.25M X 35mL = 0.1535MxV Veq. = 57mL * undergoes (c) thpon + Co? 2 Heat thos Aid Base Boise = 0.09511M (d) i) Colo Coona tho ? Cho Cooh + Nooh = -3.3x10 " + (13.3 x1 =4) - 4(3.61X153) (-3. 3X164) 2 (3.61 Xlow) x=0.26 [ht] = (2=(3+618/03) (0.26) = 9.48104 At eg. point only Naf (Salt) exists which PH = -log[ht] =pH = 3:03 is of I weak and THE Strong Base + Nauh hydrolysis . Congergate conjugate [NaF] = n = 8.75mmol 35+57) ohl PH = 7+ { pkatlog (Mar) 1: Bax is Stronger 7+1 2 13.17 Hog (0.09511) Aqueous solution would be Basic PH=8.074 (i) alty Cl + HO NHGOH + HCl 0.325g =3•61/0²M Audie because Melis Stang 1 Baginori x 0.5L e) [NO] Because only week neid and toi ile congregate salt or weak kare and its egini are cesed for Buffer Ka=Cd² + 2 = Since, Haloy is a strong and cannot be used as buffer, WA SB and seid is weak LOB SA M = m MW XV -Coo -coot осос THt COCOCH3 с ca ca C-ca 0.30 3.3x10 4 3.61x03 x>0.05 → approximation formula does not hold good. Ka= ca 2 ka-Kat=cd2 (2²+ Kad - Kazo x = -ka z J Ka ²-4 (c) (-ka)