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Vera_Pavlovna [14]
3 years ago
13

A certain metal M forms a soluble sulfate salt M2SO4. Suppose the left half cell of a galvanic cell apparatus is filled with a 5

0.0mM solution of M2SO4 and the right half cell with a 5.00M solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 20.0°C.
Which electrode will be positive?

Left

Right
Chemistry
1 answer:
VikaD [51]3 years ago
8 0

Answer:

Right

Explanation:

The given parameters are;

The soluble sulfate formed by the metal, M = M₂SO₄

In the galvanic cell formed by the metal M, we have;

The concentration of M₂SO₄ in the left half cell = 50.0 mM = 0.05 M

The concentration of M₂SO₄ in the right half cell = 5.00 M

In the galvanic cell, the metal 'M' will be dissolved into the solution with lower concentration as M²⁺ which is the left half cell, making the cell negative and the solution more concentrated

In the right half cell, the metal 'M²⁺' in the solution will be plated unto the electrode making the solution less concentrated and the electrode in the right half cell will be the positive electrode

Therefore;

The electrode which will be positive is the electrode in the right half cell.

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What is the pH if 1mL of 0.1M HCl is added to 99mL of pure water?
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Answer:

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Explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

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Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺    Kₐ₂ = 6,20x10⁻⁸; pka=7,2

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Ratio obtained is:

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As the problem said you can assume [H₂PO₄⁻] = 0,1 M and [HPO4²⁻] = 0,063M

As the amount added of HCl is 0,001 M the concentrations in equilibrium are:

H₂PO₄⁻   ⇄   HPO4²⁻ +        H⁺

0,1 M +x      0,063M -x  0,001M -x -<em>because the addition of H⁺ displaces the equilibrium to the left-</em>

Knowing the equation of equilibrium is:

K_{a} = \frac{[HPO_{4}^{2-}][H^{+}]}{[H_{2} PO_{4}^{-}]}

Replacing:

6,20x10⁻⁸ = \frac{[0,063-x][0,001-x]}{[0,1+x]}

You will obtain:

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Thus:

x = 0,063 → No physical sense

x = 0,00099990

Thus, [H⁺] in equilibrium is:

0,001 M - 0,00099990 = 1x10⁻⁷

Thus, pH of buffer after addition of 1 mL of 0,1 M HCl =

-log₁₀ [1x10⁻⁷] = 7,0

A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. In this example you can see its effect!

I hope it helps!

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