Question

what concentrations of CH3COO- and CH3COOH of those below would give a buffer with a pH of 4.11 (pKa= CH3COOH=4.84)

Br What concentrations of [CH3COO 1 and [CH1COOH1 of those below would give a buffer with a pH of 4.11? (pKa CH3COOH = 4.84). 10 a) (CH3COO]-0.15 M, [CH3COOH] = 0.10 M b) [CH3COO]-0.10 M, [CH3COOH] = 0.53 M c) [CH3COO]-0.46 M, [CH3COOH]= 0.10 M d) [CH3COO]- 0.23 M, [CH3COOH] = 0.10 M 14

Answer 1

To solve this we need Handerson - Hasselbach equation for buffer and substitute values of pH and pK_a and find value of log ratio of concentration .

Then we take antilog and find value of ratio of CH₃COO^- / CH₃COOH

Now from options we find that this ratio approximately matches to option b which gives same result

Hence option b is correct .
7 when CH3COOH and CH3600 are peegent im Solution they forme a buffer Solutions which resists Change in pH. 1. CHCOOH K L CH3600 +HT 0e Alo os Ka = [CH3C00][it] [CH₃COOH

Now According to Henderson- Hasselbach equation for this buffer pH= pkat 109 [CH3600-7 → Concentraton . [CH3 cook ] Commonion Petting values 4oll = 4.84 + 109 [CH3C00-) [CH3COOH) So loolcu = 4:11-4184 = -0073 "Генсоону - => [CH3COo ) = 0.1862 ( CH3COOH ] ICH

from given options that we find from (6) CH.COOP = 0.9 0188 = 0.10 0:53 NO.188 OH of option 1b 2 is correct.