 # A75.0 ml volume of 0.200 m nh3 (kb=1.8*10^-5) is titrated with 0.500 m hno3. calculate the ph after the addition of 25.0 ml of hno3.  pKb = - log 1.8 x 10^-5 = 4.7
moles NH3 = 0.0750 L x 0.200 M=0.0150
moles HNO3 = 0.0270 L x 0.500 M= 0.0135
the net reaction is
NH3 + H+ = NH4+
moles NH3 in excess = 0.0150 - 0.0135 =0.0015
moles NH4+ formed = 0.0135
total volume = 75.0 + 27.0 = 102.0 mL = 0.102 L
[NH3]= 0.0015/ 0.102 L=0.0147 M
[NH4+] = 0.0135/ 0.102 L = 0.132 M

pOH = pKb + log [NH4+]/ [NH3]= 4.7 + log 0.132/ 0.0147=5.65
pH = 14 - pOH = 14 - 5.65 =8.35

I hope my answer has come to your help. Thank you for posting your question here in . the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2- what are the statements

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A75.0 ml volume of 0.200 m nh3 (kb=1.8*10^-5) is titrated with 0.500 m hno3. calculate the ph after...
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