Question

What volume of 0.200 M HCl must be added to 500.0 mL of 0.250 M NH3 to have a buffer with a pH of 9.04? K, for NH4* is 5.6 x 10-10 Volume = L Submit Answer Try Another Version 2 item attempts remaining

Answer 1

When we add Hel to the NH3 buffer system, the following reaction takes place – Hatuta e rit + cio Consequently, the Henderson - Hauselballch equation for the butter system would be- i pH=pkatlog [base] [salt ] 7 pH > pika longer @ Now, we have to prepare a buffer of pH = 9.04 & pka (m+q* = -log(s6x10-0)=9.25. Using the above values, com ara @ takes the form – 904 = 925 + lang - 9.04 = 9.24 (NH3] [n++403 Now, [arit 3) = 0.54 x 0.25 mol The above moles of NH3 was available in solution, betome addition of Hel. After say we add x moles of Hel, the amount of rit, now remains will be (0.125-x) and the amount of what will be equals to 'se' moles. Hence, fotom eqo 3, we get Using the abov value atarm TK = 0.125 mol oma Hom

0.125-2 9.04 = 9.25 +- log n 2 log 0.185 * = -0.21 125-x - a a 0.21 0-125– zy n 0:125_ - -0.21 10 = 0.6166 10 ma 0.125-2 z o.6162 0.125 = 1616r =>= 0.077 moles Now, sreaction ③ takes place in inltation and an we mentioned above, that's the mo. of moles of Hel added. Hence, volume of 0.2 m Hel added= . 0.077 mad on 0.2 mot 0.387 L Hell
hence, Volume = $\large 0.387\approx 0.39L$ .