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3 years ago

Determine the number of moles NH41+ in 3 moles of (NH4)3PO4.

1 answer:

fomenos3 years ago

3 0

(NH₄)₃PO₄ is a strong electrolyte which fully dissociates into its ions. The balanced equation for the dissociation is
(NH₄)₃PO₄ → 3NH₄⁺ + PO₄³⁻

The stoichiometric ratio between (NH₄)₃PO₄ and NH₄⁺ is 1 : 3

Hence, moles of (NH₄)₃PO₄ / moles of NH₄⁺ = 1 / 3

moles of (NH₄)₃PO₄ = 3 mol

Hence, moles of NH₄⁺ = moles of (NH₄)₃PO₄ x 3
= 3 mol x 3
= 9 mol

Hence, 3 moles of (NH₄)₃PO₄ give 9 moles of NH₄⁺.

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3 years ago

9. Write a balanced nuclear equation for the following: The isotope Strontium-90 decays by Q-

Sergio039 [100]

Answer: $_{38}^{90}\textrm{Sr}\rightarrow _{36}^{86}\textrm{Kr}+_2^4\textrm{He}$

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3 years ago

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3 years ago

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3 years ago

For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ

Komok [63]

Answer:

Check the explanation

Explanation:

Answer – Given, $H_3PO_4$ acid and there are three Ka values

$K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}$

The transformation of $H_2PO_4- (aq) to HPO_4^2-(aq)$ is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

= 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

= 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

= -log 6.2x10-8

= 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

= 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

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3 years ago