Question

I dont not expect anyone to create the spreadsheet, however I am totally lost on the calculations and values that must be added, so any help on that would be greatly appreciated!

INTRODUCTION An acid-base titration is a procedure for carrying out a neutralization reaction between an acid solution and a base solution by the controlled addition (from a burette) of one of the solutions, known as the titrant, to the other solution. For such a titration, a graph of solution pH versus volume of titrant added is known as a titration curve. Such curves may be used to determine the equivalence point of the titration and to estimate the ionization constant (K, or K) if a weak acid or weak base is titrated, The general shape of all titration curves is similar. For most of the titration the pH changes only gradually with titrant volume. The exception to this is a very narrow volume region from just before to just after the equivalence point. In this region there is a large change in pH with titrant volume. See the figure below for an example of a typical curve. Using "HA" to represent any weak acid, the net ionic equation for the reaction occurring during the titration of HA with a strong base is HA(aq) + OH-(aq) - H200) + A-(aq) (1) 10 > 10 Titration of a Weak Acid with a Strong Bano Page 1 of 13

Answer 1

Equivalence point. * Volume of Acid sample = 25.0 mL Concentration of acid = .0.1030 (M) 1 of NaOH = 0.1188 (M) - ml volume of NaOH at 25x 0.1030 equivalence point := 0.1188 = 21.6 75 m2

C D E F G H I K L M N O P Q R S T B 1 Volume of NaOH pH 1 3.337 4.07 4.43 Volume of NaOH Vs PH 5.2 5.38 Equivalence point:- 21.675mL at pH 8.3 6.06 6.42 6.89 8.2 9.89 10.35 11.4 11.65 11.86 10.8375 ml at pH 5.13 12.3 35.9 401 12.47 12.51 12.62 B 20 25 30 35 40 45

• Volume of NaOH at equivalence point = 21.675 mL

Volume of NaOH at half equivalence point = 10.83 mL

pH at half equivalence point = 5.13

• At half equivalence point , pH = pKa

Therefore pKa = 5.13 => Ka = 10^(-5.13) = 7.413*10^(-6) at the reaction time .