Question

10:COMPARISON OF BUFFERED AND UNBUFFERED SYSTEMS To Sindents: This labit hranih alat date wahisand calculations of buffers and pH, there is no date. Your instructor will assign a specific buffer for or instructor will assign a specific buffer for you to make in the lab. "You will then be asked to perform serveral calculations based on your migue buffer assignment. way bandurite or be your answers directly on this party for submission to a Blackboard link Remember to show all your work in order to receive credit Purpose Statement: he purpose of this experiment is to make a buffer solution when given a Ph and concentration of a buffer through a series of calculations. pH mL Buffer Assignment concentration (M) 0.050 M 6.40 100.00 Determine the Mass of Each component Recall that buffers are formed from conjugate acid/base pairs. Using the information given about your assigned buffer, determine how much of each component (acid and base) you will need in order to prepare it in the lab. (This will require a system of equations because there are two "unknowns.") Note: the conjugate acid in this case is H,PO, and the conjugate base is HPO.. Using equation (3) in your lab manual, solve for the ratio of (moles B/moles A). Phphat log Mus) 6.40: 1.11+ log() 10 .15 3. From the ratio, solve for the moles B is forms of moles A. (Hint: you are not solving for the actual number of moles B at this point, you are only solving for the relationship of B to A.) (B/A = 15) A + B = .15A Using your concentration and volume, solve equation (1) in the lab manual for (buffer). Sem - .050 M = Sest .100 Stot= .005 mol

I need help with questions 4 to 7. I have solved 1 to 3 as reference. Thank you.

Answer 1

4 molos of A + mellos of B = 0.005 mol. & B = 0.15 A moles of A = 0oog mot a numele 1:15 moles o B 20.00 -0.00433 - 0.00067 mol. moles B = 0.00433 x 0.15 mol = 0.00067 mola 5 mass of conjugate aced a 0.00433 mol x 137399 g/mol. = 0.5979. mars of base = 0.000 67 mol x 268.19/mol. = 0.1809 HIPO4 +HU H₂PO4 + co Or000 335 0.403x101.00 0,0043Z -OSO 2165 mol mol = 0.000103 mol. 0:00 232 mol. 0.002268 mol papa2 tlor (Halog 7.21+ loq (.000 232 ) 001002268 p = 6.22