Consider a buffer made by adding 52.3 g of (CH3),NH2I to 250.0mL of 1.42 M (CH3)¿NH (Kb = 5.4 x 10-4) What is the pH of this buffer?

Chemistry & Chemical Reactivity
9th Edition
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter17: Principles Of Chemical Reactivity: Other Aspects Of Aqueous Equilibria
Section17.2: Controlling Ph: Buffer Solutions
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Consider a buffer made by
adding 52.3 g of (CH3),NH21 to
250.0 mL of 1.42 M (CH3)¿NH
(Kb = 5.4 x 10-4)
What is the pH of this buffer?
1
3.
6.
C
7.
8.
6.
LO
4.
Transcribed Image Text:Consider a buffer made by adding 52.3 g of (CH3),NH21 to 250.0 mL of 1.42 M (CH3)¿NH (Kb = 5.4 x 10-4) What is the pH of this buffer? 1 3. 6. C 7. 8. 6. LO 4.
Expert Solution
Step 1

The concentration or Molarity of salt is calculated by -

M = w×1000/m×V

Here,

M = Molarity of Salt, (CH3)2NHI = ?

w = mass of solute = 52.3 g

m = molar of (CH3)2NHI = 12×2+1×6+14+1+127= 24+6+14+1+127 = 172 g/mol

V = volume = 250 mL

 

Then,

M = 52.3×1000/172×250

M = 1.2163 M (CH3)2NHI

Or

M = 1.22 M

 

Thus , concentration of dimethyl Ammonium iodide is 1.22 M.

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