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blagie [28]
3 years ago
5

Write the net ionic equation for the precipitation of silver phosphate from aqueous solution:________

Chemistry
1 answer:
SVEN [57.7K]3 years ago
5 0

Answer: 3Ag⁺ (aq) + PO₄³⁻ (aq) → Ag₃PO₄ (s)

Explanation:

The question requires us to write down the net ionic equation for the precipitation of silver phosphate from aqueous solution:________

Note that we we told to:

(Use the solubility rules provided in the OWL Preparation Page to determine the solubility of compounds.)

Answer:

From the solubility rule,

Phosphate (PO₄³⁻) and Carbonates are insoluble except for NH₄⁺ and Group1A compounds.

According to the solubility rule, the precipitation of silver phosphates and that of the net ionic equations is given thus;

3AgNO³ (aq) + K₃PO₄ (aq) → Ag₃PO₄ (s) + 3KNO₃ (aq)

while the Net ionic equation is given;

3Ag⁺ (aq) + PO₄³⁻ (aq) → Ag₃PO₄ (s)

cheers i hope this helped !!

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Answer: Option (d) is the correct answer.

Explanation:

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3 years ago
Write the Henderson-Hasselbalch equation for a propanoic acid solution ( CH3CH2CO2H , pKa=4.874 ) using the symbols HA and A− ,
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Answer:

a) [A⁻]/[HA] = 0.227

b) [A⁻]/[HA] = 0.991

c) [A⁻]/[HA] = 2.667

Explanation:

In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:

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pH = pka + Log [A⁻]/[HA]

pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]

  • (a)

4.23 = 4.874 + Log [A⁻]/[HA]

-0.644 = Log [A⁻]/[HA]

10^{-0.644} = [A⁻]/[HA]

0.227 = [A⁻]/[HA]

  • (b)

4.87 = 4.874 + Log [A⁻]/[HA]

-0.004 = Log [A⁻]/[HA]

10^{-0.004} = [A⁻]/[HA]

0.991 = [A⁻]/[HA]

  • (c)

5.30 = 4.874 + Log [A⁻]/[HA]

0.426 = Log [A⁻]/[HA]

10^{0.426} = [A⁻]/[HA]

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