Question

inn? 1. Formic acid, (K= 1.8x10) a WA, and sodium formate, its' CB, are mixed together in a 250.0 mL volume of 0.350 M formic acid and 0.455 M sodium formate- to form a buffer solution. a. If 1.00 mL of 1.00 M HCI is added to 250.0 mL of DI water, what is the new pH? What is the pH if the 250.0 mL of solution was only 0.350 M HCOH (no formate)? b. c. What is the pH of the original mixed solution? What is the pH of solution (c) above after adding 1.00 mL of 1.00 M HCI? d.

Answer 1

[ 4 coon] 20.3519 ; [ncove] =0.455 from Henderson Hanselbalch ele. ph of buffer sohn 11 ph= 3-77 they of 0.455 pn = 3-88 [ncr]; = 109.2 [ue] } g 103 mole Ent new = 0.25 otte = 4x10m. ph= leg [ar - 3- leegy 2 2.397 6 Hoon Hif HC 0.3514-0.35 0.50 0.354 poke, Cun Turis) [newn] 0.350 xossal -3-77 oss (1-x) = 0.358 Ji- aal FITSEE Ltd., FIEE House, 29. A. Kali Sami Sanu

0.35 CX USS &= (4.85 X104) = 2.2010-2 [he] = ossa = 0.35X2.2010 7.78103 p" - eg (2-4 X1073) 0.88 - 3 - = 2.11 Ipn

ha 4e ao Hal - de to 10 mole 2 4coon to 0-3 8x250 nede & 118:15 mind + 20 0-3 SX250 • 113.251 min 113 75-2 87-572 due to connes on effect que Simu ph - ples & by Sal? since nat is very small and me olution is buffer so it will do alnust do not disturb the ph f sofa so pha 3.88 /1034875