Question

You are given 2.0 L of a buffer solution that contains 0.220 M acid (pka- 3.50) and 0.200 M of conjugate base. A. What is the pH of the buffer? B. What volume of 0.500 M solution of HCI would you need to add to make the pH of the buffer solution 3.00?

Answer 1

Buffer mixture which has to Resist That miature the ph change quantity small a high tendency upon addition of very of strong acid or strong buffer soln soln and Meature, base is called Buffer mixture. Acidic . basic Buffer Buffer labially mixture of it is the basically it is the sniature of weak acid + salt og weak lase t its heeak acid with salt with strong acid eg NM QOH į NHAC. strong d. CH 2600 HA CH₃200 Na eg.

Acidic Buffer eg. CH₃Cool $ & rolar . Colg COONa c molar . CHg cool 2 CHgcool + Ht. tao с. to ega a lo-a) Cta ao an kas ce+aa] [Aaj & C1-2) M al ctaa) [l-aj 又くく then hoe ineglect it Ka. alexga) - xso -1 ka zac La Ka ka then, CHE] a ad Now, taking - log a ka e - Long (HT) - log aka с

PH pka f log & salt] (acid ] pH = pka + ls ty-2) (t2 3 case. let 2 rol of Naon is added. 2 < 1 . Сн, Соон + NaOH — CH₃Coona + Ноо. t2 O 2 , 8. Eat 2 - 2 О. 4+ 2 2 ең а pRq 1 Log salt log Coad). Рң Ź eka + log C4+2 ) ta-2)

comes to this question volume of leaffer 2.OL tacid 0.220 M РКа of acid = 3-50 I conjugate base or salt ) = 0.2009 Now, This is the case of acidic be buffer PH pka t log [ salt tacid } . so PH !) 3.50 t + log to.200] {0.220 РЧ 3.45

B ) what volume of 0.500 a M Solution of Hel whould whould you need to add to make the PH of the buffer solution 3.00?

final moles of conjugate base = of conjugate base = 0.1 -0.50 final moles of acid 0.84 to.sv Now, this the acidie buffer so / [ Salt or conjugate PH = pką t log! base [acid] so PH pka 1) + log! long ( moles of conjugate base) (moles of acid ) given deta's outs but all the las volume cancel