Identify the oxidizing agent and the reducing agent for 4Li(s) + O_2 (g) to 2Li_2O(s).

1 answer:

7 0

The reaction is:

<span>4Li(s) + O2 (g) = 2Li+ + 2O-2(s).

The oxidizing agent is the one that is being reduced which is oxygen where the charge changed from neutral to -2 while the reducing agent is the on being oxidized which is lithium where the charge change from neutral to +1.</span>

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Phosphoric acid (h3po4) is tribasic, with pka's of 2.14, 6.86, and 12.4. The charge of the ionic form that predominates at ph 3.

Annette [7]

Correct answer is $H_2PO_4^-$ .

Phosphoric acid is a polyprotic acid having 3 acidic hyrdogen therefore it will have 3 pka values.

The equations for the release of acidic hydrogen can be written as:

$\\H_3PO_4\rightleftharpoons H_2PO_4^-+H^+\text{ pka=2.14}\\H_2PO_4^-\rightleftharpoons HPO_4^2^-+H^+\text{ pka=6.86}\\HPO_4^2^-\rightleftharpoons PO_4^3^-+H^+\text{ pka=12.4}$

From the pka values we can judge the idea of pH as using Henderson-Hasselbalch Equation, we get the relation between pH and pka.

$HA\rightarrow H^-+A^-$

Using the following equation, relation of pH and pka is

$pH=pka+log\frac{[A^-]}{HA}$

Using this equation, we can find that the equation having pka= 2. 14 is closest to the pH=3.2 so the ionic form in this equation will be dominant at the same pH.

Therefore at pH=3.2 the ionic form $H_2PO_4^-$ of $H_3PO_4$ is dominant.