Question

A solution is prepared by dissolving 0.23 mol of formic acid and 0.27 mol of sodium formate in water sufficient to yield 1.00 L of solution. The addition of 0.05 mol of NaOH to this buffer solution causes the pH to increase slightly. The pH does not increase drastically because the NaOH reacts with the present in the buffer solution. The K, of formic acid is 1.8 x 10+ A formic acid B. sodium formate C water D. sodium E. strong acid

Answer 1

Исоон + HCOONa 0. 23 mol 0.27 mol Icoone koooo tur 0.05 mol Naon HCOONa coot Nat of of Add" Before add" of base pH = pka + log [ncoona] After add" of bace [ncoon] base will react with ut ions and will give ressant of ait left and pH = pkat log Cucoond +x] mere sale Incoon x = ho off of on o] HCOOH + Noon HCOONa+H₂O Hence formic acid reacts with Naola I Acidosis means increase in concentration kidney and lunge, which lowers the pH band the patient will be unable to breathe. When the kidney and lungs are releasing excess of con pH of blood which maintains at O 7-35 goes down. Acidosis, the ph is below 7-35. Ill Buffer of (weak acid and its salt) or (weak base and its salt) O us s a solution As hf is a weak acid and its salt Naf HF = ut +fo Naf , Hf will make a buffer solution.

IV -Coon = + 4* Hoenderson - Hasselbalch Equation pH = pks + log Lisodium benzoate ] [ benzoic acid Concha pka -logka Volume 0-51 no. of moles pH = 4.2 logo 0.05/ = 3.2 pH = 4.2+(-1) pH = 3.20 2+ 2- Calon), Ca +2lon) CaCO, Ca ²+ + Gou S 2S 3 Ksp = 453 Ksp = S? s- Vase s-3 kse 4 S = 0.011 M S = 5-196 X/05 m 2. Caso, Calt + Sou? 28 S S Caz (PO4)2 =36&* + 2PO43. 35 Rsp = 10855 S: 7.13 X 10 M Ksp = 52 s-3x 103 m M Molar Solubility (s) is highest for Calon), zik, 0.011 M Cuzcoon Cu₂c000 + H® pH = pka log [Chz cona pH = 4.74 + log ( 0.80 [CH₂coon] - 5.04 ph=

-pka + log MAJ VL Let barbituric acid be HA НА > и* + А teo C - + с (1-) ca ca Ka = ca ² 9. & x/0-5 1 - О•2 2 = 4.8 x10-5 1 - а * = 49 х 10° (1-а ) = 0 - 0 2 Ги+] : с. - 0 : 2 х 0-0 2 О, ОИМ pH = -log(0.04) ри = : 30 VIL ИА + лаби » АNa + И, О А ph = ри = ач •o | 4-он | 2 (40 р/ 24-01 loghi 모 ч 0° Tx си, соои + NaOH Д-30-15 Os » СИ, (ooNa + И, о б: 15 РИ = 4-71 + Дар (o-15) (о: 15 ) ри = 4-74
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