Question

Consider the titration of 80.0 mL of 0.0200 M CH3NH2 (a weak base; Kb = 0.000440) with 0.100 M HNO3. Calculate the pH after the following volumes of titrant have been added:

(a) 0.0 mL

pH =

(b) 4.0 mL

pH =

(c) 8.0 mL

pH =

(d) 12.0 mL

pH =

(e) 16.0 mL

pH =

(f) 24.0 mL p

H =

Answer 1

la) and Ho is shewn belon The Reaction between CH3 NH C4NH Initial o.0200M fnal O.O2008 O.020001- O.02004 Degnee of dissodatim dissoiatiom Cmstant OO00440 Base X o.02 pod O.0200 O.O20002 ke weak Be Kb O.00044 1 o.0200 O.0200 O.1483 7O.020 OPON00XO.14832 2-96X103 он -lol2 26x1) poH 2.53 14-PO PH 14-2-53 11-4 Amsne PH

Cddition of HNo 4.omL NO OF O.100MX 4-omL add ed = mamales o HNo O-4m.male O.02001X 80.0mL mmale o CHaNH added NO 1-6 m.mde No of m.mde of HNo cidded No of mamale of salt feamed 1 o.4 m-mole 1-6-0-4)m-moe NO ofm-mole of CH3N left 1-3m.male Pka of CH&NH= 1o-63 PKb of CHH PH=14-3.3e +1g 3.32 CCHhM LCNH Handenson -Hasselbal Equahon 14-(3-38 -+13 1-2 m-mole/v =Total Volume of &olutim V 14 -3.38 t-4) 11. Ansne PH addition ot HNO NO ot m.mole of HNo, added=o. 100MX8-omL B.omL -o.8m.mok No of m.male of cHN No of m.mole of salt foamed NO of mmde of CH3N udded 1-6 mmal o.8m.mole Left -6-0-8)m-mk 1 O.8male

14- PH O.mmale 3.38 + 109 1 4- PH Amane lo-62 No of mmole of HNo Cdded O-10CMX 12-omL 1-2m-mole 17 No of m.mole of CHaNH No of m.male af satt foome d t-6m.mole N00f mmMe of HNO added 11 1-2mmole 1 No of m.male of LL6-1-21m.mele CHANH Left O-4m.mule PH=14-pk+ N Handens Harselbalch e quation 20malelv 4m-mele 3-38+09 14- O. PH 10.14 Ansoes