Question

3. Calculate the pH at the following points for the titration of 25.0 mL of 0.100 M formic acid (Ka = 1.80 x 10-4) with 0.100 M NaOH: VNaOH = 0.00 mL, 15.00 mL, 25.00 mL, 40.00 mL. Draw a graph of pH vs. VNAOH. (30 points) (b) Buffer capacity can be thought of as how well a solution resists changes in pH after a strong base/acid is added. A buffer is most effective to resisting pH changes when what is true, explain. (5 points)

Answer 1

fornic acid 3. Titration of asml of 0.1M HCOOH (Ka 21,88104) with o. lm Nowy in Unaoh = Ome. - HCROM Cags - Heco (aq) + Hilag, Altzo o.IM Kaol.8810-4 At t= tee Ool-a 01 X eventolioxley 22x18x150 [Mt] = 404,243815 3m 1 PM = -log[MT] = -log(432438153) = 2:37]

c) VNOOH = 15ml from Newt (laushing reactant) HCOOH + lor Hс + H, o Cars Yay Kee - (Ka)нсон cay I Kw 2 хо, Is Xool - М to 40 2 1.8 x 1.9 10-14 21. 9х1о •Г- ) чо — fқ чо Ester 9. – 4 х 4. А), чо ) lis о — 2 ), хі. % xxx 2. Г 1. x to'° = х - (бн ) о» 033 x to". Сой) рои - - Сой ) = looя. | РК = /ч - грок 3.92 (ii) Unadh a dome нсън + он => Acco +н, °, lе, 1» Proto 2rxool ихо, | от 2 О. з д•1 м 2.

o, | - 2 1, 8х10 24 о» 0257 х 10-1», + 3 Гой) t = vo so 2 + xo /> ... < xo- pon - - -у (ой) , г. ?? | e4 г 14 - rоf 8 2 2 1 Unool - Yome, Hewllaq! + oH cas, Hccolaql etha, kes 2rxo :) Ab Иохо» * de to Y68x llo ге (r Xo-1, 2. Хо/ Ке Come é toxola или 2 – - poka-log (15) = 1.64. \ r / Гри? /4 -roH • 19-1-64 - 2, 3 .

- 18:22 - quivalence Prints - - - - - - - - - - Naoh (3) Buffer capacity of a Buffer Slection tells us about the strength of Buffer, how well it resist the change in pH upen open addition of acid/ Base for monimum Buffer capacity. 1 [conjugate Base }e (weak acid] (for a codic Puffer) [p4 pha [pH = pke] Condations of . [wryugate and] =( weak base] for Basic Buffer 20m 2 (pon o pki] melde Buffon