Question

NICO What mass of Na3PO4 must be added to 69.4 mL of 0.296 M HCl to obtain a buffer with a pH of 7.86? Ka (H3PO4) = 7.5 x 10-3 K(H2PO4) = 6.2 x 10-8 K(H,PO.) = 3.6 x 10-13 Mass =

Answer 1

Given 0.296M Hd = 69.4 ml pH = 7.86 The reaction is . Nazpont Hel - Naztpou tnad Naz Hpoyt Hel > Malla poy + lad Total mass of Hel = 0.296x0.0694 -0.0205 mol .: Total mags of Hd = 0.0205 mol Let us assume moles o Nazpon x and mole of Hol remain after reduction = 0.0205-7 mole of Nagtipon present = x mol . Let us assume y mole of Nagtipon is converted

2 into Nath pop then Y = 0.0205-7 mole of Mazh pou remain = x-10.0205-8) 3 22-0.0205 -male-of waz #poy semain = 22-0.0205 p" - plasty ( 93-0005] 0.0205-7 >> 7.86 7.2+ 22-0.0205 0.0205-? => 7.86-7.2 22 -0.0205 0:0 205-7 => 786-72 song --> 0.667 dog ( 2010 0900) => 0.66=1 (27-0.0205 0:0 205-7 = > 0.66 2a-0.0205 - 0.0205-7 lo 22-0.0205 = 4:570 0.0205-x => => 29-0.00205 = 4.5 7010.0205-2) 29-0.00205 = 0.09368 - 4.570.00

> 224 + 4.5702 -0.00205-009368 = 0 6.57 % - 0.09573 -0. => 6.572 0.09573 7 0.09573 6.57 2=0.01457 mol ) mass of Naz poy = 0.01457X164 = 2.3896 Na3 pou mase of Nazpon = 2.3896 Nazpou) 2 please date me.