Question

I could really use help for question 1 and 2! Thank you in advance :)

MATERIALS AND EQUIPMENT . 1 - pH probe w/LabQuest . 1 - 100 ml beaker . 1 - stir bar • 1 - stir rod • 3 - medium beaker • 2 - 30 mL beaker CHEMICALS • sodium carbonate, NaHCO3 • 0,1 M hydrochloric acid, HCI • 0.85% lactic acid • dry ice • ammonium chloride, NHACI assigned acid assigned conjugate base • 6 M hydrochloric acid, HCI • 6 M sodium hydroxide, NaOH PROCEDURE & DATA HAZARDS Use caution with all chemicals. 6 M NaOH and HCI can cause severe burns. PART A: BLOOD BUFFERS Add 50 mL DI water to a 100 ml beaker. Place a stir bar in the beaker and set the spin speed to a very low rate just enough so that it is spinning. Remove a pH probe from the storage solution and rinse with DI water. Clamp the pH meter into the beaker, being sure that the pH probe does not come in contact with the stir bar. Record the pH. Add -1.25 g NaHCO3 (record exactly to the beaker, and wait for the NaHCO, to dissolve. You may turn up the stir bar slightly to facilitate dissolving, then immediately reduce the speed. Record the pH. If it is not close to 8.2 your pH probe may need to be calibrated. Add - 15 mL 0.1 M HCl (record exactly to the beaker. Record the pH. Question 1. Based on your procedure, how did you form this buffer system? Which gas was produced? Use appropriate chemical reactions in your answer. Question 2. How does this pH compare to the pH of blood? Is there more carbonic acid or bicarbonate present? Explain.

Answer 1

The answer is attach below

DI water volume = 50ml Acetat Naficaz adding = 10259 after Add = 15mL of colm HCl ANSWER T. We form the buffer of H₂CO3 / Nahcez in which H₂CO3 act as acid part of buffer and Nasicą act as conjugate bare part of beeffes. Note: No My Cz from by adding to the HCin Nallcoz → CO2 g) produced appropriate chemical reactions NaHCO3 (22) + HCl en Na Cl + H₂CO3 €2) (unstable) # Hy (3 en is unstable, so bh CO₂ ED - H₂ Opt CO2(g) This gay is produced. Answer:21 Now calculation of pol of Buſker. Cue assume that H₂ CO₂ is not going decomposition) mass of NaHCO 3 ) = 1.25g Molar mass of Nanco = 84.007 glomel mole of Nalle g = 1.25% 84.007 good = 0.01486 mol Total volume of solution - 50ml weeter + 15m2 nci = 65m2 molarity of Nallcg 0.0 1488 mol = 0.2289 M 65mL

Male of HCl = volume of HCL x melarity of HCS Mole of HC1 = 15ml * 0.1 m 1.5 mel Mclarity of Haci in Solection - Dicle cok Plci Volume of solutions = 1.5 rýmist 65th Molosity of HC ir suden 0.023077 mei NaHCO3 + HClan Nacht H ₂ CO 3 6 Net Ionic equation - Hooge the 7 H₂CO3 (4) Before HCI 0.2289M Addihin 0.2289M - ) +0.023077ms! After HCl addition +0.02989 mol 0-23017M/ Ateqully +0.001871 0.02219M we assume little amount de composed still - H₂CO3 amount is large , while Nancoz is zere. from this procedure you not get buffer because conjugate base is not present or completely neutralize .

pll this =) 8.2 C ginen) blood ph=) 7.1 both are almost basic, it butter is from than pH is near to pka of Nanco lcg, that is near to 6.3 , but in acheal your solution is not forming get better, - from above calculation we see that H₂CO3 (Carbonic acid) is more amount present than NäH(03_( bicarbonate present). H₂CO3 = 0.0 225 mol while Nahcoz a zero Explanation is above calculation. selony