Question

Buffer Practice 1.) You wish to prepare 750 mL of a buffer solution of pH = 7.00 and you decide to use NaH2PO4 as the acid, K2 = 6.2 x 10. You have on hand a 0.175 M solution of this acid. Kai = 7.5 x 103, Ka2 = 6.2 x 10-8, K3 = 4.2 x 10-13 a.) Calculate the pH of the NaH2PO4 solution. b.) Calculate the grams of Na2HPO4 that need to be added to 750 mL of the NaH2PO, in order to produce a buffer of pH = 7. c.) You have on hand also, a 0.120 M solution of Na2HPO4. How many mL of the NaH2PO, and how many mL of that Na2HPO4 solutions should be mixed in order to produce 750 mL of a pH = 7 buffer? d.) You have a 0.15 M solution of KOH. How many mL of the KOH solution should be added to the 750 ml of the NaH2PO4 solution to produce a pH = 7 buffer? What is the total volume after the addition? e.) How many mL of the NaH2PO4 solution and the KOH solution are required to produce 750mL of a pH = 7 buffer? f.) In c.) you computed the volumes of NaH2PO4 and NaH2PO, required to make a pH = 7.00 buffer. Accidentally, you added the calculated volume of NaH2PO4, but misread the label. You aytually added that volume of 0.120 M of H3POinstead. What is the pH and is it a buffer?

Answer 1

Buffer solution of pH = 7.00 having 7.50 aving 750ML volume it Acid Na H₂ POy of Kan = 6.2x100 .2x10 8 in it = of acid 0.1751 ution of concentration Kai : 7.5Y10- 3 Ka = 6.2 x 10-8 shoot ng tot valed smo Kaz= 4.28 10-13 l oldot t el 1 Ha Ha pou – Mad Hapoy (Bolt pubg boopH H₂ Pou + NaOH na e 7 prisibin to old, o ut kail to con t=t cox 1 H₂ Pou t = t x-y Kar Hout & Ht ly; pou 3- + H+ + skaz HPO 2 y-z e x (x+y+z) x2 Ka, [H] [ He pou [H₂ pou] (0.175-x) 0.175 a y y (147] ovodo (x +y+z ) H. Polo y a --- day to do H2 pon dodo pou [poy & I [+] ) zxx Sz x (x+y+z kazy и Роу y-z pou buty

хі с О. 11 утссе 9 ka, E У ТСәe +y+7 92 | О» | 7 7 7 >> Ч77 2 { | x Ка, - ух Ка, Као - ?? Као ч олио - ГОН Ka, c. 15. have - 5 xo - . Як рибоох 18-01x & della 2 х = 1, 6 хоз хр. 17 iddin 1.3128x10-3 - да, зму: у ху). mg = 0 - оз озм х - - [н'] 2 3.63х10 - Гн*] - 3.63x102 м C - To E)+ 1 = PHI та н, РОy ) = 9 - 1973 63 - - уч

on (a) PH of Na H₂PO4 & 1.44 to (6) hm of achive PH Nag Hpoy vequired ct solution 750ml is 7 Ka = y = 6.2 x 10 - 8 오 4, 2010- ZX 3.63x100 = H4.2 X10-13 6.2x10-8 8 4.2 x 10-13 x 6.2x10-8 ZTTO 3.63 x 10-2 -19 .17 X 10CM : y = [H Pou] = 0.2410-8 m moles of Nag HPO = 6.2x10 - 8 x 750 1000 ci 3465 X10-8 ma se Hрои : ч. 65х10 - &х чи = 655 x 10-6 gm