Question

Part B: A 30.0-mL volume of 0.50 M CH3COOH (Ka=1.8×10?5) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH at 25 ?C. Express the pH numerically.

MasteringChemistry: ASSIGNMENT #6 (Chapter 16)-Google Chrome https://session.masteringchemistry.com/myct/itemView?assignmentProblem ID=59386 148 CHEM 101 (M03) Help | Close NMENT #6 Titration of Weak Acid with Strong Base Resources Y previous | 19 of 25 | next » ± Titration of Weak Acid with Strong Base A certain weak acid, HA with a Ka value of 5.61 × 10 6 , is titrated with NaOH A titration involves adding a reactant of known quantity to a solution of an another reactant while monitoring the equilibrium concentrations. This allows one to determine the concentration of the second reactant. The equation for the reaction of a generic weak acid HA with a strong base is Part A A solution is made by titrating 9.00 mmol (millimoles) of HA and 3.00 mmol of the strong base. What is the resulting pH? Express the pH numerically to two decimal places. HA(aq) + OH (aq)→A (aq) +H2O(1) pH= 4.95 Submit Hints My Answers Give Up Review Pa Correct Part B More strong base is added until the equivalence t is reached. What is the pH of this solution at the equivalence point if the total volume is 33.0 mL? Express the pH numerically to two decimal places Submit Hints My Answers Give Up Review Part Continue tO Ask me anything 1:34 PM 3/27/2016

Answer 1

In this neutralization reaction, 0.5 M CH COOH titrate with 0.5 M NaOH to produce sodium acetate and water with equal volumes. Total volume of solution30 mL + 30 mL = 60 mL , At equal amount of moles and volume doubled, the concentrati on of acetate becom es half of its intial i.е., 0.25 M The acid dissociation constant (Ka) of acetic acid is 1.8×10-5 The base dissociation cons tan t (K^)is as follows: K. 1x10-4 K, 1.8x103 -5.55x1010 At the end point, sodium acetate i s hydrolyzed to produce hydroxidions CH,COO-(aq) + H20(i) 근 CH3COOH (aq) + OH-(aq) initi al Change E quilibriunm 0.25M 0.25-х b 0.25-x 5.55×10-10- 0.25-x X2 = 5.55×10-10 × 0.25 (Neglect the small value of x x =| OH- |=1.178×10-5 pOHl0g OH log 1.178x10- - 4.93 pH 14-4.93 = 9.071 Therefore, the pH at the end of titration is 9.07