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Vsevolod [243]
4 years ago
8

When aqueous solutions of chromium(II) iodide and sodium phosphate are combined, solid chromium(II) phosphate and a solution of

sodium iodide are formed. The net ionic equation for this reaction is:
Chemistry
1 answer:
Georgia [21]4 years ago
4 0

Answer:

3 Cr²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Cr₃(PO₄)₂(s)

Explanation:

When aqueous solutions of chromium(II) iodide and sodium phosphate are combined, solid chromium(II) phosphate and a solution of sodium iodide are formed. The molecular equation is:

3 CrI₂(aq) + 2 Na₃PO₄(aq) ⇄ Cr₃(PO₄)₂(s) + 6 NaI(aq)

The full ionic equation includes all the ions and the molecular species.

3 Cr²⁺(aq) + 6 I⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇄ Cr₃(PO₄)₂(s) + 6 Na⁺(aq) +         6 I⁻(aq)

The net ionic equation includes only the ions that participate in the reaction and the molecular species.

3 Cr²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Cr₃(PO₄)₂(s)

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A 0.2 M carboxylic acid (RCOOH) has a Ka = 1.66x10-6. What is the pH of this solution? Enter to 2 decimal places.
maria [59]

Answer:

3.24

Explanation:

The dissociation equation for the carboxylic acid can be represented as follows:

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We can use an ICE table to get the value of the concentration of the hydrogen ion. ICE stands for initial, change and equilibrium.

RCOOH RCOO- H+

Initial 0.2 0.0. 0.0

Change -x +x. +x

Equilibrium 0.2-x. x. x

We can now find the value of x as follows:

Ka = [RCOO-][H+]/[RCOOH]

(1.66* 10^-6) = (x * x)/(0.2-x)

(1.66 * 10^-6) (0.2-x) = x^2

x^2 = (3.32* 10^-7) - (1.66*10^-6)x

x^2 + (1.66 * 10^-6)x - (3.32* 10^-7) = 0

Solving the quadratic equation to get x:

x = 0.0005753650094369094 or - 0.0005753650094369094

As concentration cannot be negative, we discard the negative answer

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pH = -log(0.0005753650094369094)

pH = 3.24

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3 years ago
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