How much of 0.3 M NH4OH should be mixed with 30 mL of 0.2 M solution of NH4Cl to give buffer solution of pH 8.65? (Given: textpK textb of textNH text3 = 4.75 )

Question

How much of 0.3 M NH4OH should be mixed with 30 mL of 0.2 M solution of NH4Cl to give buffer solution of pH 8.65? (Given: textpK textb of textNH text3 = 4.75 ) (A) 2 mL (B) 3 mL (C) 5 mL (D) 7 mL

Tardigrade Admin · Accepted Answer

It is a basic buffer pH=8.65 , pOH = 14 - 8.65 = 5.35 pOH=pKb+log([. overset⊕N H4 ]./[. N H4 O H ].) 5.35 = 4.75 + log ([. overset⊕N H4 ]./[. N H4 O H ].) ([. overset⊕N H4 ]./[. N H4 O H ].)= antilog (5.35 - 4.75) = antilog (0.6) ≈4.0 So ([. overset⊕N H4 ]./[. N H4 O H ].)=(30 × 0.2/0.3 × V)≈4.0 So V = 5 mL.

Q. How much of 0.3 M NH4​OH should be mixed with 30 mL of 0.2 M solution of NH4​Cl to give buffer solution of pH 8.65? (Given: pKb​ of NH3​=4.75 )

2 mL

3 mL

5 mL

7 mL

It is a basic buffer pH=8.65 , pOH = 14 - 8.65 = 5.35 pOH=pKb​+log[NH4​OH][N⊕H4​]​ 5.35 = 4.75 + log [NH4​OH][N⊕H4​]​ [NH4​OH][N⊕H4​]​= antilog (5.35 - 4.75) = antilog (0.6) ≈4.0 So [NH4​OH][N⊕H4​]​=0.3×V30×0.2​≈4.0 So V = 5 mL.

Q. How much of 0.3 M $N H_{4} O H$ should be mixed with 30 mL of 0.2 M solution of $N H_{4} Cl$ to give buffer solution of pH 8.65?
(Given: $pK_{b}$ of $NH_{3} = 4.75$ )