Question

A 250 mL buffer solution is prepared by using x g
of NH4CI and V mL of 1 M NaOH solution.
At pH = 9.6, the concentration of NH4+ ion in final
solution is 0.1 M. The value of x + V is (Kb of
NH4OH = 2 x 10-5)​

Answer 1

Answer:

Its a basic buffer pH=9;pOH=5

pOH=pK

b

+log[

(Base)

(Salt)

]

5=−log(108×10

−5

)+log[

NH

4

OH

NH

4

Cl

]

5=4.745+log[NH

4

Cl]

no. of Moles of NH

4

Cl added=1.8

Answer 2

Final answer: 54.01  

Given that: We are given  250 mL buffer solution is prepared by using x g of NH₄CI and V mL of 1 M NaOH solution, At pH = 9.6, [NH₄⁺] in final solution = 0.1 M,  $K_{b}$ of NH₄OH= 2×10⁻⁵.

To find: We have to find x + V.

Explanation:

• pOH of buffer solution containing weak base and its salt is given by,

         pOH = p$K_{b}$ + ㏒ $\frac{[Salt]}{[Base]}$

        pOH = 14 - pH

                  = 14 - 9.6

                  = 4.4

• p$K_{b}$ = -㏒$K_{b}$     [ $K_{b}$ = 2×10⁻⁵]

               = -㏒(2×10⁻⁵) = 4.7

• [Salt] = [NH4+] = 0.1 M, [Base] = [NH₄OH] = ?

• By substituting these values to the equation of pOH,

            4.4 = 4.7 + ㏒ $\frac{[0.1]}{[NH4OH]}$

           (4.4 - 4.7) = ㏒ $\frac{[0.1]}{[NH4OH]}$

           anti㏒[-0.3] = $\frac{[0.1]}{[NH4OH]}$

          [NH₄OH] = 0.2

• NH₄CI + NaOH → NH₄OH +NaCl

Now number of moles of NH₄OH is produced = number of moles of NaOH is reacted

        i.e., [NH4OH] = [NAOH]

      [NH₄OH] = [(V * 1) / 250] = 0.2

From this V = 50 ml.

• Total number of moles of NH₄CI = Number of moles of NH₄⁺ formed + Number of moles of NH₄⁺ left in solution

        = (50 * 1) + (0.1 * 250) = 75 * 10⁻³ mole

• x = (75 * 10⁻³) * molecular weight of NH₄CI

           = (75 * 10⁻³) * 53.5

           = 4.01

• x + V = 4.01 + 50 = 54.01

To know more about the concept please go through the links

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