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Gekata [30.6K]
3 years ago
10

In the test for Cl1- and SO4 2- , explain why the addition of acid will dissolve or decompose other silver salts and barium salt

s, which otherwise might Interfere
Chemistry
1 answer:
myrzilka [38]3 years ago
3 0

Strong acids can dissolve the salts of weak acid. When we consider the different salts of silver:

Salts of silver with the conjugate bases of a weak acid are soluble in strong acidic solutions. Some of these salts are:

2Ag^{+}(aq)+CO_{3}^{2-}(aq) --->Ag_{2}CO_{3}(s)

Ag^{+}(aq)+C_{2}H_{3}O_{2}^{-}(aq)-->AgC_{2}H_{3}O_{2}(aq)

2Ag^{+}(aq)+SO_{3}^{2-}(aq) --->Ag_{2}SO_{3}(s)

Salts of silver with the conjugate bases of a strong acid are not affected by change in pH:

Ag^{+}(aq)+Cl^{-}(aq) --->AgCl(s)

2Ag^{+}(aq)+SO_{4}^{2-}(aq) --->Ag_{2}SO_{4}(s)

These two salts with Chloride and sulfate ions are not soluble in acidic solutions as the salts of silver with the conjugate bases of a strong acid are not soluble in acidic solutions, they remain unaffected by any change in pH.

So for salts of Ag and Ba with the conjugate bases of a weak acid, solubility is increased upon the addition of an acid. So, the interference from the ions of weak acids can be removed by decreasing the pH.

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When performing your experiment, you add your first drop of oleic acid/benzene solution, and a lens forms immediately. what is w
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7 0
3 years ago
Read 2 more answers
Calculate the cell potential at 25oC under the following nonstandard conditions: 2MnO4-(aq) + 3Cu(s) + 8H+(aq) ⟶ 2MnO2(s) + 3Cu2
baherus [9]

Answer:

1.346 v

Explanation:

1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:

(oxidation) Cu_{(s)} →Cu^{2+}_{(aq)} +2e E°=0.337 v

(reduction) MnO_{4 (aq)} + 3e + 4H^{+}_{(aq)}→MnO_{2 (aq)}+2H_{2}O E°=1.679 v

(overall) 2MnO_{4 (aq)}+3Cu_{(s)}+8H^{+}_{(aq)}→3Cu^{2+}_{(aq)}+2MnO_{2 (aq)}+4H_{2}O E°=1.342 v

2) Nernst Equation

Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:

E=E^{0} -\frac{RT}{nF}Ln\frac{[red]}{[ox]}

Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.

The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give

E=1.342 -\frac{298.15*8.314}{6*96500}Ln\frac{[.66]}{[1.69]}=1.346

E=1.346

5 0
3 years ago
How many seconds would it take to deposit 17.3 g of ag (atomic mass = 107.87) from a solution of agno3 using a current of 10.00
Brums [2.3K]
Data Given:

Time = t = ?

Current = I = 10 A

Faradays Constant = F = 96500

Chemical equivalent = e = 107.86/1 = 107.86 g

Amount Deposited = W = 17.3 g

Solution:
             According to Faraday's Law,

                                          W  =  I t e / F

Solving for t,

                                          t  =  W F / I e
Putting values,
                                          t  =  (17.3 g × 96500) ÷ (10 A × 107.86 g)

                                          t  =  1547.79 s

                                          t  = 1.54 × 10³ s
5 0
3 years ago
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