*My main instruction: 1. You can find in attachment a file, named (Chemical Analysis -module one ), in this file contain a questions about introduction chemistry, introduction spectroscopy and IR...

Chemical Analysis Activity:
 Introduction (equili
ium):
Question 1
Write each answer with the co
ect number of digits:
a) 3.021 + 8.99 = 12.01
d) 0.0302 ÷ (2.1143 × 10
-3
) =14.3
) 12.7 – 1.83 = 10.9
e) log (2.2 × 10
-18
) = -17.66
g) 10
-4.555
= 2.79x10-
5

c) 6.345 × 2.2 = 14x10
1

f) antilog (–2.224) = 5.97x10
-3
Question 2
Find the absolute and percent relative uncertainty and express each answer with a reasonable number
of significant figures:
a) 3.4 (±0.2) + 2.6 (±0.1) =
     
1 2
2 2 2 2
1 2
7
3.4 0.2 2.6 0.1 6.0 e
e 0.2 and e 0.1
e e e 0.2 0.1 0.224
Absolute uncertainty is 0.2 and relative uncertainty is 3. %
Answer is 6.0 0.2 or 6.
    
 
      
 70 3. %

) [3.4 (±0.2) × 10-8] ÷ [2.6 (±0.1) × 103]
   
 
8 3
1 2
2 2 2 2
1 2
11
3.4 0.2 10 2.6 0.1 10
e 0.2 and e 0.1
e e e 0.2 0.1 0.224

1.308 0.092 x 10


          
 
      
 

c) 3.4 (±0.2) ÷ 2.6 (±0.1)
     
       
       
2 2 2 2
1 2
3.4 0.2 2.6 0.1 1.308 e
3.4 0.2 /2.6 0.1 3.4 5.88% /2.6 3.85%
%e %e %e 5.88 3.85 7.03
Absolute uncertainty is 0.092and relative uncertainty is 7.0%
Answer: 1.308
    
    
      
 00.092 or 1.30 7. %

d) [3.4 (±0.2) – 2.6 (±0.1)] × 3.4 (±0.2) =
     
    2
3.4 0.2 – 2.6 0.1 0.8 0.224 0.8 28%
0.8 28% 3.4 5.88% 2.7 29% 2.72 0.78
     
      

Question 3
a) A solution is prepared by dissolving 0.2222 (±0.0002) g of KIO3 [FM 214.0010 ±0.0009] in 50.00
(±0.05) mL. Find the molarity and its uncertainty with an appropriate number of significant figures.
      33 3 3
6 8
Molarity 0.222 2 0.000 2 g of KIO * mol of KIO / 214.001 0 0.000 9 g of KIO * 50.00 0.05 x 10 L
0.020 76 0.000 02 M
   
 

) Would the answer be affected significantly if the reagent were only 99.9% pure?
Molarity is the number of moles of solute per liter of solution. As number of mass depends on purity
of the substance therefore, if the purity is only 99.9%, the molarity will be different. So, the answer will be
affected if the purity of the reagent changes.
Question 4
Using activity coefficients co
ectly, calculate the pH of
a) 0.050 M HBr
Answer:
As it is a strong electrolyte, activity is same as that of concentration.
3The pH is by definition the negative logarithm of the H O .
[ 3 ] [ ] 0.050
log[ 3 ] log[0.050] 1.3
H O HBr M
pH H O
  
  
     

) 0.050 M NaOH
Answer:
As it is a strong electrolyte, activity is same as that of concentration.
0.050
log[ ] log[0.050] 1.3
14
14 1.3 12.7
NaOH M
pOH OH
pH pOH
pH

     
 
  
Question 5
Find the pH and fraction of dissociation (α) of a 0.010 0 M solution of the weak acid HA with
Ka = 1.00 × 10
-4
.
Answer:
Since the Ka value is not low and the concentration is high, it should be possible to use the ICE
method.
HA H+ + A
Initial 0.010
Change -x +x +x
Equili

2
2 -6 -3
ium (0.010 - x) x x
1.00 x 10-4 = [H+][A-]/[HA] = x /(0.010 - x)
If we assume x
0.100, this becomes x = 1.00 x 10 or x = 1.00 x 10 M.
pH = 3.000.
- fraction of dissociation
=[A-]/[HA] /x

  -3 -3(0.010 ) 1.00 x 10 / (1 1.00 x 10 )...