Question

6.)Part B

A 64.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 32.0 mL of KOH.

Express the pH numerically.

7.) Part C

Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8×10−5) with 0.20 M HNO3. Calculate the pH after addition of 50.0 mL of the titrant.

Express the pH numerically.

8.) Part D

A 30.0-mL volume of 0.50 M CH3COOH (Ka=1.8×10−5) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH.

Express the pH numerically.

Answer 1

Concentration of HBr= 0.25 M Volume of HBr 64.0 mlL moles ofHBr= 0.25 M × 64.0mL= 0.016 mol Concentration of KOH 0.50 M Volume of KOH32.0 mL moles of HBr0.50 Mx32.0 mL0.016 mol Write the reaction as follows: Since moles of HBr is equal to moels of KOH The pH of the soluti on will be equal to 7 Therefore, the pH of the solution is 7 Write the reaction as follows: Concentrati on of NH0.20 M Volume of NH3- 50.0mL moles ofNH3-0.20M×50.0mL= 0.01 mol Concentrati on of HNO3-0.20 M Volume of HNO3 - 50.0mL moles of HNO3-0.2011x 50.0mL= 0.01 mol moles of NH,'-0.01 mol NH4- +7 0.01mol 100 mL = 0.1M Draw the ICT table for the ionis ati on of NH as follows I(M): 0.1 0 C(M):-x E(M): 0.1-xxx Given K 1.8x103 K,-1.0×10-14 = 5.56×10-⑩ K, 1.8×10 x 5.56x10-10- x- 7.45x106 M H+]= 7.45x10-6 M pH-log H =-log(7.45x10-6) -5.13 Hence the required pHis 5.13

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(D)