Question

1- 30.00 mL of sodium sulfite solution are titrated by 45.78 mL of 0.000983 M potassium dichromate in the presence of sulfuric acid, generating sulfate and chromium (III) ions. Determine the balanced net ionic equation for the titration, and calculate the molarity of the sodium sulfite solution.

2- A voltaic cell is constructed at 25˚C using half cell couples of Fe3+/Fe2+ with 1.00M/0.100 M concentrations and MnO4 - /Mn2+ at 1.00x10-2 M/1.00x10-4 M concentrations in an acidic solution of 1x10-3 M hydronium. Find the electrode potential for each half cell, and the overall cell potential.

Answer #1

1.

Net ionic equation between dichromate and sulphite solutions is,

Cr_{2}O_{7}^{2-} (aq.) + 3
SO_{3}^{2-} (aq.) + 8 H^{+} (aq.) 2
Cr^{3+} (aq.) + 3 SO_{4}^{2-} (aq.) + 4
H_{2}O (l)

Useful formula is,

M_{1} V_{1} / n_{1} = M_{2} V2 /
n_{2}

Given data,

Molarity of idchromate, M_{1} = 0.000983 M

Volume of dichromate, V_{1} = 45.78 mL

Moles of dichromate, n_{1} = 1 mol

Molarity of sulphite solution, M_{2} = ?

Volume of sulphite solution, V_{2} = 30.00 mL

Moles of sulphite, n_{2} = 3 mol

Therefore,

0.000983 * 45.78 / 1 = M_{2} * 30.00 / 3

M_{2} = Molarity of sulphite solution = 0.00450 M

A buffer solution is prepared by mixing 65.5 mL of 0.768 M
sodium hydrogen sulfite with 25.3 mL of 0.0706 M sodium
sulfite.
Calculate the pH (to two decimal places) of this solution.
pKa hydrogen sulfite ion is
7.19
Assume the 5% approximation is valid and that the volumes are
additive.
Calculate the pH (to two decimal places) of the buffer solution
after the addition of 0.0173 g of sodium sulfite
(Na2SO3) to the buffer solution above.
Assume 5% approximation...

You are titrating 100.0 mL of 0.0200 M Fe3+ in 1 M HClO4 with a
titrant solution of 0.100 M Cu+ to give Fe2+ and Cu2+ (E0 Cu2+/Cu+=
0.161V; E0 Fe3+/Fe2+= 0.767V).
What is the cell potential after pouring 10.0mL of titrant?

A solution containing the following was prepared at 25°C: 0.18 M
Pb2 , 1.5 × 10-6 M Pb4 , 1.5 × 10-6 M Mn2 , 0.18 M MnO4–, and 0.86
M HNO3. For this solution, the following balanced reduction
half-reactions and overall net reaction can occur.
5[Pb4+ + 2e- --> Pb2+]
E°= 1.690 V
2[MnO4- + 8H+ + 5e-
--> Mn2+ + 4H2O] E°=1.507
V
_______________________________________________________
5Pb4+ +
2Mn2+ + 8H2O <-->
5Pb2+ + 2MnO4- + 16H+
A) Determine E°cell,...

Using
0.200 ml H2O2
10 mL H2SO4 (aq) 1 M
Solution titrated with KMnO4(aq) 0.20M
In the reaction of the redox pairs MnO4- | Mn2+ and H2O2 | O2.
Show reaction scheme and calculate the formal concentration
c(H2O2)

Using
0.200 ml H2O2
10 mL H2SO4 (aq) 1 M
Solution titrated with KMnO4(aq) 0.20M
In the reaction of the redox pairs MnO4- | Mn2+ and H2O2 | O2.
Show reaction scheme and calculate the formal concentration
c(H2O2)
35% h2o2 and density=1.126 g/ml

1. A 100.0-mL sample of 0.500 M sodium hydroxide is titrated
with 0.100 M nitric acid. Calculate the pH before the titration
begins.
2. A 1.0-L buffer solution contains 0.100 mol HCN and 0.100 mol
LiCN. The value of Ka for HCN is 4.9 x 10-10. Because the initial
amounts of acid and conjugate base are equal, the pH of the buffer
is equal to pKa = -log (4.9 x 10-10) = 9.31. Calculate the new pH
after the addition...

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