Problem 17.46-Enhanced-with Feedback Constants I Periodic Table Consider the tit
ID: 1025510 • Letter: P
Question
Problem 17.46-Enhanced-with Feedback Constants I Periodic Table Consider the titration of 30.0 mL of 0.050 M NHs with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added. You may want to reference (Pages 729-739) Section 17.3 while completing this problem. Then, the pH is calculated from the concentration of OH- as follows Part B 20.0 mL Express your answer to two decimal places. pH- Submit Previous Answers Request Answer Incorrect; Try Again; 3 attempts remainingExplanation / Answer
NH3 -----BASE
HCl -------ACID
A.
when HCl = 0ml added
pOH = -log [OH-] = -log 0.05 = 1.3
pH = 14 - pOH = 12.7
B.
when HCl = 20ml added
let x ml of NH3 reacted
x*0.05 = .025* 20 => x = 10
Therefore the amount of free NH3 will get decreaesd.
Conc becomes 2/3 rd of initial
pH = 14 - log[ 0.05*.667] = 12.523
C.
let x ml of NH3 reacted
x*0.05 = .025*59 => x = 29.5
thus, pH = 14- log [0.05*1/60] = 10.92
D
. let amount of hcl used be x ml
x*0.05 = x *0.025
=> x = 60 ml
But HCl taken = 60 ml
ph = 7
e. let amount of hcl used be x ml
x*0.05 = x *0.025
=> x = 60 ml
But HCl taken = 61 ml
ph = -log [0.025/61] = 3.387
f.
5 ml of HCl will be left
thereefore, ph = -log 0.025*5/65 = 2.716