Question

Problem Set 17 CHEM 1252 November 18, 2019 Textbook Reading Assignment 16.5-8 Additional Practice Problems: 16.51,55-58,63,73,75,79,81-82,89,91,93,95,97-98,105,123,127 1. A buffer called "This is widely used in biochemistry. The neutral) compound Tris itself is a weak base, So we'll notate it as "B." The pk, of its conjugate acid (HB) is 8.07. The overall goal of this problem is to calculate the pH change of a Tris buffer upon addition of some strong base. (a) Assume that you have a pH 8.40 Tris buffer with a total concentration (B) + THB'1-0.0500 M. Calculate the (B/THB ratio. Use this ratio and the total concentration to find the concentration of each species (b) How many mmol (millimoles) of B and mmol of HB are in 50.0 mL of this buffer? (c) Write the net ionic equation for the reaction that happens when OH is added to a B/HB buffer. Why can we safely assume that it goes to completion"--what is the equilibrium constant for this reaction? (d) Calculate mmol of OH in 1.00 mL of 0.200 M NaOH. (e) Using your answers from parts b, c, and d, calculate mmol of B and mmol of HB* after the addition of 1.00 mL of 0.200 M NaOH to 50.0 mL of the pH 8.40 Tris buffer. (0) Find the new (B/THB ratio and calculate the resulting pH. [Shortcut: remember that buffer pH does not change upon moderate dilution, so total volume doesn't matter herell

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Answer 1

LB For Tris buffer, Henderson Hasselbalch equation is PH=pka & og [B] - [BHT] a) pka ²8.07 PH=8.40 Substituting in egr 0 8.40 = 8.07 & wg [B] 0.332 wg (B] (But) [B]/(BHT) = 100.33 - 2014 Now, [B] + [BHT] = 0.0500M [B] - 0.0500 - (BHPJ 0.0 500 - (BHT) 22.14 CBHJ [BH] = 10.0159 M) [B] = 0.0500-00159 = 10.0341M] b) Millimoes & Molarity x volume (ml) Millimoles of B = 0.0341 M X 50 ml a 1.705 Millimol Millimoles of BH t = 0.0159 MX 50 ml - 0.795 millimoles

c) Ot will react with BHT - H (aq) +BHt (ag) - Blag) thole) It goes to completion because it is an acid- bax realtion resulting in neutralization to form water. Equilibrium constant will be very high. a) mmol of ot - Molarity of Napk x -0.200M X 100 ml = 0.200 mmol e) As from ionic equation in part c) BHt reacts with or to form B. so, millimoles of B = 1.705 +0.200 1905 millinole Millimoles of BHT= 0.1950.200 = nad millimole. 0.595

& New [B] 2 1.905 & 3.20 [BH] 0.595 New pH can be calculated from eau ① PH ₂8.07 + g 3.20 = 18.58] New pH after addition of Nad - 8.58