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lara31 [8.8K]
3 years ago
13

xFormula1" title="H_2PO_4^-(aq) \rightarrow H^+(aq) + HPO_4^{2-}(aq)" alt="H_2PO_4^-(aq) \rightarrow H^+(aq) + HPO_4^{2-}(aq)" align="absmiddle" class="latex-formula">
Find the pH of this buffer after 0.069 mol of OH- was added to 1.00 l of this buffer containing 0.165 M is H_2PO_4^- and 0.594 HPO_4^{2-} (Ka₁ = 7.2 x 10⁻³; Ka₂ = 6.1 x 10⁻⁸; Ka₃ = 4.2 x 10⁻¹³) assume no volume change due to the addition of OH⁻.
Chemistry
1 answer:
klasskru [66]3 years ago
4 0

Answer:

The pH of the buffer solution = 8.05

Explanation:

Using the Henderson - Hasselbalch equation;

pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]

where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21

Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)

[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M

[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M

Therefore,

pH = 7.21 + log (0.663 / 0.096)

pH = 7.21 + 0.84

pH = 8.05

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Reduction reactions are the reactions in which loss of oxygen takes place.

For a given reaction:

2Na(s)+2H_2O(l)\rightarrow H_2(g)+2NaOH(aq.)

Sodium is getting oxidized because there is an addition of reaction with that element.

Hydrogen is getting reduced because there is a removal of oxygen with that element.

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G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu
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The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄Ag^{+} + Br^{-1}.

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Or, 5.0×10^{-13} = S^{2} (the given value of solubility product of AgBr is 5.0×10^{-13} and the charge of the both ions are same).

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<h3>Avogadro's hypothesis </h3>

From Avogadro's hypothesis,

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2.5×10²⁴ atoms = (2.5×10²⁴ × 2) / 6.02×10²³

2.5×10²⁴ atoms = 8.31 g of H

<h3>How to determine the mass of water </h3>
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Learn more about Avogadro's number:

brainly.com/question/26141731

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