Question

23. What is the amount of ammonium chloride that has to be dissolved in 190 mL of 0.189 M NH3(Kb= 1.8×10‒5) to make it a buffer with a pH of 8.96, in grams?

24. A mixture contains 0.250 M benzoic acid, a monoprotic acid (Ka = 6.28 × 10‒5), and 0.400 M sodium benzoate. How many mL of a HCl solution whose pH is 0.523 should be added to 500 mL of this buffer to change its pH from what it is to pH = 4.20?

Answer 1

Answer of the question 1 = 19.84 gram

Answer of the question 2 = 125.418 mL

& Chemical equilibrium exists in a solution of ammona is K6 NHGOH NHut tot ko? b [NHGOH] [Nhut [OH-] Now let [NH+] = [OH-]-S e [NHGOH] bet art to 1.8x10-5 - 82 f ebboat rotalog [NHGOH ? So =) 822 1.8x105 x 0,189 a 3-402x10-6 2) s-10845 x103 M = TNHt] [ok] = initiale Now letas consider amof NH gel is added in the solution of pt 2 8.06 or poltz (14-8.06) 25.04 ne So pky a poh – log (+1.845x103) 2 4.744 25.06 -log (at 1-845x103) 0.189 x+10845 x 10-3 21.077 0.189 2) 19.849 12 x 0.371 M (0.37 1x 5315) gram So 19.84 gram of Nhul to be added.

Molar A pH of the Hel solution = 0.523 So & concentration of ution in this solution is a -0.523 32 10 Ma 0.299 M IN Letas consider x ml of this Hol 0.299 M Hal solution to be added to the buffer. are 30 number of moles of Att ins in this solution is a xx0.299 moles. a (x 2,90 810-4) moles holes of benzoic acid in sooml buffere 0.250m Moses of benzoate in soond buffer 0.40x500 1000 1000 20.125 moles 10000.0 moles Now If new PH 2 412 then (benzoate pka z pH = log [herzoic acidy pka z PH (benzoie 0.20 € 2x2,0mg x 10" 3) - log (6.23 310-5) -2- by one 0.125 + x x 2.999 =) 462 = up-lz (0.20 - xx2.90x184) (0.125+ xx2:99810-4 ) =) 0.125 + xx 2,94810- 4 0.20 - xx 2.99x10-4 xz 125.418 ml