Chemistry, 24.12.2019 22:31 maddo36

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.

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Answer from: jjmo

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

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Answer from: Quest

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-

Answer from: Quest

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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