Consider the titration of 70.0 mL of 0.0300 M CH3NH2 (a weak base; Kb = 0.000440) with...

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Chemistry

Consider the titration of 70.0 mL of 0.0300 MCH3NH2 (a weak base; Kb =0.000440) with 0.100 M HNO3. Calculate the pH after thefollowing volumes of titrant have been added:


(a) 0.0 mL

pH =  

(b) 5.3 mL

pH =  



(c) 10.5 mL

pH =  



(d) 15.8 mL

pH =  



(e) 21.0 mL


pH =  



(f) 27.3 mL

pH =  

Answer & Explanation Solved by verified expert
4.3 Ratings (852 Votes)
The trick to solve this kind of exercise is to know where you are in titration curve It is well know that in the equivalence point MaVa MbVb 70 0030 01 Va Va 21 mL So at 21 mL is the equivalence point At 0 mL CH3NH2 H2O CH3NH3 OH i 003 0 0 e 003x x x Kb x2 003x but Kb is a small value so the value of x therefore 003x 003 x 440x104 00312 x OH 363x103 M pOH log363x103 pOH 244 pH 14244 1156 b at 53 mL In this case we have now the acid so    See Answer
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