50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pKb of ammonia solution is 4.75, the pH of the mixture will be :

Question

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pKb of ammonia solution is 4.75, the pH of the mixture will be : (A) 3.75 (B) 4.75 (C) 8.25 (D) 9.25

Tardigrade Admin · Accepted Answer

Number of millimoles of NH 3 = 50 mL × 0.2 M =10 mmol Number of millimoles of HCl = 25 mL × 0.2 M = 5 mmol 5 mmol NH 3 will react with 5 mmol HCl to form 5 mmol NH 4 Cl 10-5 = 5 mmol NH 3 will remain. pOH = pK b - log ([ NH 4 Cl ]/[ NH 3]) pOH =4.75- log (5 / V /5 / V ) pOH =4.75 pH =14- pOH =14-4.75=9.25

Q. 50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pKb​ of ammonia solution is 4.75, the pH of the mixture will be :

3.75

4.75

8.25

9.25

Number of millimoles of NH3​=50mL×0.2M=10mmol Number of millimoles of HCl = 25mL×0.2M=5mmol 5mmolNH3​ will react with 5mmolHCl to form 5mmolNH4​Cl 10−5=5mmolNH3​will remain. pOH=pKb​−log[NH3​][NH4​Cl]​ pOH=4.75−log5/V5/V​ pOH=4.75 pH=14−pOH=14−4.75=9.25

Q. 50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If $p K_{b}$ of ammonia solution is 4.75, the pH of the mixture will be :