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Appendix_II_Practice+Problems_Key_LZM__v+F10

# Appendix_II_Practice+Problems_Key_LZM__v+F10 - MCB 120L...

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MCB 120L AII-1 Answers to Practice Problems: Experiment 1 1.1. pH = pKa + log (A/HA) A + HA = total concentration of acetic : acetate acid 4.06 = 4.76 + log (A/HA) 0.2 = A/HA solve for A & HA by ratio analysis OR algebraically with 2 equations : 2 unknowns ratio: __ algebraically:___________________________________ A = (0.2/1.2)(0.2M) = 0.0333 M A + HA = 0.2 M & 0.2 = A/HA HA = (1/1.2)(0.2M) = 0.167 M 0.2 HA = A, then by substitution, 0.2 HA + HA = 0.2 M HA = 0.167 M, then 0.167 M + A = 0.2 M, A = 0.0334 M grams of NaAcetate = (0.0333 M)(0.1 L)(136 g/mol) = 0.453g milliliters of acetic acid = (0.167 M)(0.1 L)/17.6 N = 0.000949L, or 0.949ml V 1 C 1 = V 2 C 2 1.2a. 5.06 = 4.76 + log (A/HA) A + HA = 0.2 M HA = 0.0667 M (algebraic solution) 2 = A/HA 2 HA + HA = 0.2 M A = 0.133 M (0.133 M)(0.1 L)(136 g/mol) = 1.81 g NaAcetate (0.0667 M)(0.1 L)/17.6 N = 3.79 x 10 -4 L, or 0.379 ml acetic acid 1.2b. (0.050 M)(0.1 L)/ 0.2 M = 0.025 L, or 25 ml of 0.2 M acetic acid buffer at pH5.06 with H 2 O bring final volume to 100 ml 1.3 To make 0.050 M acetate buffer at pH 5.5, first recognize that the acetate available (CH 3 COOK) is the potassium salt, meaning, it is all in the base form. The base form must be titrated with HCl to form the conjugate acetic acid in the ratio giving pH 5.5. grams of CH 3 COOK required = (0.25 L)(0.050 M)(98.15 g/mol) = 1.227 g ~ 1.23 g [H + ] = [HA] to achieve the A/HA ratio to reach a pH of 5.5; ie titrate the acetate (conjugate base) to achieve a pH of 5.5. 5.5 = 4.76 + log ([A]/ [HA]) A/ HA = 5.495 [HA] = [H + ] = (1/ 6.495)(0.050 M) = 0.007698 ~ 0.0077 M HCl (0.0077 M)(0.25 L) / (1.0 N HCl) = 0.00192 L ~ 1.92 mls of 1.0 N HCl 1.4a. Na 2 CO 3 , 106 g/mol NaHCO 3 , 84 g/mol HCO 3 - H + + CO 3 - 10.38 = 10.38 + log (A/HA) A = HA the pKa is the pH when A = HA A + HA = 0.2 M carbonate buffer = 0.1 M CO 3 - + 0.1 M HCO 3 - Na 2 CO 3 (0.1 M)(0.1 L)(106 g/mol) = 1.06 g NaHCO 3 (0.1 M)(0.1 L)(84 g/mol) = 0.84 g 1.4b. (0.08 M)(0.1 L) / 0.2 M = 0.04 L, or 40 ml or 0.2 M carbonate buffer at pH 10.38 with H 2 O bring final volume to 100 ml

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MCB 120L AII-2 1.5. Tris in crystalline form is a base, often purchased as TrisBase, which is the conjugate Tris-base for the Tris buffer. The quantity of HCl needed is equal to the quantity of Tris (base) titrated to form the conjugate Tris-acid. before HCL is added, A = total quantity of Tris buffer (0.2M)(0.1L)(121g/mol) = 2.42g Tris-base 8.71 = 8.21 + log (A/HA) = 8.21 + log (Tris-base/Tris-acid) A + HA = 0.2M A = 3.16HA 3.16HA + HA = 0.2M HA = 0.048M, which equals the quantity of HCl added (0.048M)(0.1L)/11.7M = 0.00041L, or 0.41ml of HCL 1.6. Before titrating Tris to the desired pH, Tris is entirely in the conjugate-base form, A. (0.2M Tris)(121g/mol)(0.1L) = 2.42g at pH 8.45, 8.45 = 8.21 + log A/HA and 0.2M Tris = A + HA solve for HA, the [HCl] needed to titrate A and bring the pH to 8.45: HA = 0.073M volume of HCl added: (0.073M)(100mls)/11.7N = 0.62ml, or 620μl 1.7a. Glycine in H 2 O: + H 3 NCH 2 COO pKa’s = 9.91 and 2.36 (respectively) 17b. pH = (9.91 + 2.36) / 2 = 6.135 ~ 6.1 1.7c. Addition of 50 ml 0.1M KOH to 50 ml 0.2 M glycine will result in the titration of the amine (ie the removal of protons), the carboxyl group will remain fully unprotonated. pH = 9.91 + log {(0.0 M NH 2 + 0.05 M KOH) / (0.1 M NH 3 + – 0.05 M KOH)} = 9.91 1.7d. Consulting Table 1.1, page 1-4, the pKa will decrease, so the pH also will decrease. 1.8a. Addition of 50 ml 0.1M HCl to 50 ml 0.2 M glycine will result in the titration of the carboxyl (ie the addition of protons), the amine group will remain fully protonated.
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