Question

How many moles of CH₃NH₃Cl need to be added to 200.0 mL of a 0.500 M solution of CH₃NH₂ (Kb for CH₃NH₂ is 4.4 × 10⁻⁴) to make a buffer with a pH of 9.70?

Answer 1

Answer:

how much moles of m solution need

Answer 2

Given:

1. Volume of solution (v) = 200 mL = 0.2 L
2. Molarity of Base (M) = 0.5 M
3. Kb of base = 4.4×10⁻⁴
4. pH of required buffer solution = 9.70

To find:

Moles of CH₃NH₃Cl required to be added.

Solution:

• Let [salt]  be the concentration of CH₃NH₃Cl and [base] be the concentration of CH₃NH₂
• The pKb of the base = -log(Kb) = 3.30
• The pOH of the buffer solution = 14-pH = 4.30
• For a basic buffer solution, pOH = pKb + log([salt]/[base])

     ⇒4.30 = 3.30 + log([salt]/[base])

     ⇒log([salt]/[base]) = 1

     ⇒[salt] = 10×[base] = 10×0.5 = 5M

• The number of moles of CH₃NH₃Cl required is equal to 1 ([salt]×v)

Answer:

The number moles of CH₃NH₃Cl required to be added is equal to one.