What is the Kb of the base PO−34 given that a 0.48M solution of the base has a pH of 12.70? The equation described by the Kb value is PO−34(aq)+H2O(l)⇌HPO−24(aq)+OH−(aq)   Report your answer with two significant figures.

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What is the Kb of the base PO−34 given that a 0.48M solution of the base has a pH of 12.70? The equation described by the Kb value is

PO−34(aq)+H2O(l)⇌HPO−24(aq)+OH−(aq)

 

Report your answer with two significant figures.

**Determination of the Base Dissociation Constant (K<sub>b</sub>) for PO<sub>4</sub><sup>3−</sup>**

**Problem Statement:**
What is the K<sub>b</sub> of the base PO<sub>4</sub><sup>3−</sup> given that a 0.48 M solution of the base has a pH of 12.70?

**Chemical Equation:**
The equation described by the K<sub>b</sub> value is:

PO<sub>4</sub><sup>3−</sup> (aq) + H<sub>2</sub>O (l) ⇌ HPO<sub>4</sub><sup>2−</sup> (aq) + OH<sup>−</sup> (aq)
Transcribed Image Text:**Determination of the Base Dissociation Constant (K<sub>b</sub>) for PO<sub>4</sub><sup>3−</sup>** **Problem Statement:** What is the K<sub>b</sub> of the base PO<sub>4</sub><sup>3−</sup> given that a 0.48 M solution of the base has a pH of 12.70? **Chemical Equation:** The equation described by the K<sub>b</sub> value is: PO<sub>4</sub><sup>3−</sup> (aq) + H<sub>2</sub>O (l) ⇌ HPO<sub>4</sub><sup>2−</sup> (aq) + OH<sup>−</sup> (aq)
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