Appendix_II_Practice+Problems_Key_LZM__v+F10

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 36

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online