1 ) 2.52×10 2 L sample of a solution of Cu + requires 3.28×10 2 L of 0.133 M KMn
ID: 996765 • Letter: 1
Question
1 ) 2.52×102 L sample of a solution of Cu +requires 3.28×102 L of 0.133 M KMnO4solution to reach the equivalence point. The products of the reaction are Cu2+ and Mn2+.
2) An electrochemical cell is based on the following two half-reactions:
Ox: Sn(s)Sn2+(aq, 1.50 M )+2e
Red:ClO2(g, 0.105 atm )+eClO2(aq, 1.50 M)
Compute the cell potential at 25 C.
3 ) A 115.0 mL sample of a solution that is 2.6×103 M in AgNO3 is mixed with a 225.0 mL sample of a solution that is 0.14 M in NaCN.
After the solution reaches equilibrium, what concentration of Ag+(aq) remains?
4)55.0 mL of 0.17 M HCHO2 with 75.0 mL of 0.11 M NaCHO2
5 ) 130.0 mL of 0.10 M NH3 with 250.0 mL of 0.10 M NH4Cl
CALCULATE THE PH FOR 4 ND 5
Explanation / Answer
1) 1 mole of KMnO4 reacts with 5 moles of Cu+
moles of KMnO4 = 0.133 M x 3.28 x 10^-2 L = 4.3524 x 10^-3 mols
moles of Cu+ = 5 x 4.3524 x 10^-3 mols = 0.021812 mols
concentration of Cu+ in solution = 0.021812 mols/2.52 x 10^-2 L = 0.865 M
2) Eo = 0.954 - (-0.1375) = 1.0915 V
Ecell = Eo - 0.0592/n logK
= 1.0915 - 0.0592/2 log(1.5^2 x 1.5/0.105^2)
= 1.018 V
3) moles of Ag+ = 2.6 x 10^-3 M x 115 ml = 0.299 mmol
moles of CN- = 0.14 M x 225 ml = 31.5 mmol
Ag+ + 2CN- <==> [Ag(CN)2]-
moles of [Ag(CN)2]- formed = 0.299 mmol
[Ag(CN)2]- = 0.299 mmol/340 ml = 8.8 x 10^-4 M
remaining CN- = 31.5 - 2 x 0.299 = 31.201 mmol
[CN-] = 31.201 mmol/340 ml = 0.092 M
Kf = [Ag(CN)2]-/[Ag+][CN-]^2
5.6 x 10^18 = 8.8 x 10^-4/0.092^2 x [Ag+]
[Ag+] concentration = 1.85 x 10^-20 M remains
4) pH = pKa + log(base/acid)
= 3.75 + log[(0.11 M x 75 ml/130 ml)/(0.17 M x 55 ml/130 ml)]
= 3.69
5) pH = pKa + log(base/acid)
= 9.26 + log[(0.1 M x 130 ml/380 ml)/(0.1 M x 250 ml/380 ml)]
= 8.976
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