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3 years ago

Oxalic acid, H2C2O4, reacts

Chemistry

1 answer:

Dominik [7]3 years ago

6 0

Answer:0.005M

Explanation:

First deduce the oxidation and reduction half equations and from that obtain the balanced redox reaction equation. From that, the number of moles of reacting species are seen from the stoichiometry of the reaction from which the number of moles of oxalate is obtained and substituted to obtain the molar concentration of oxalate.

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Answer:

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Explanation:

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3 years ago

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EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.

Back to this question. $[\text{H}^{+}]$ is given. 25 °C implies that $K_w = 10^{-14}$ . As a result,

$\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}$ .

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DochEvi [55]

Answer: 40.1%

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