You know oxygen is the limiting reactant (since it says there is excess hydrogen). So, use stoichiometry based on the given number of oxygen moles:
The equation described by the kb value is 5.21 x
.
The potential of Hydrogen is what pH is formally known as. The negative logarithm of the concentration of H+ ions is known as pH. Thus, the definition of pH as the amount of hydrogen is provided. The hydrogen ion concentration in a solution is described by the pH scale, which also serves as a gauge for the solution's acidity or basicity.
Assuming PO₄³⁻ (the phosphate anion).
PO₄³- + H₂O ==> HPO₄²⁻ + OH⁻
Kb = [HPO₄²⁻][OH⁻] / [PO₄³⁻]
We can find the [OH⁻] from the pH of 12.70.
pH + pOH = 14
14.0 - 12.7 = 1.3 = pOH
[OH] = 1x
[OH-] = 5.0x
M
[HPO₄²⁻] = 5.0x
M
Kb = (5.0x
)2 / 0.48
= 2.5x
/ 0.48
Kb = 5.21 x 
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Answer is: dissolve 74,9 grams CuSO₄·5H₂O in one liter volumetric flask.
V(CuSO₄·5H₂O) = 1 L.
c(CuSO₄·5H₂O) = 0,30 mol/L.
n(CuSO₄·5H₂O) = V(CuSO₄·5H₂O) · c(CuSO₄·5H₂O) .
n(CuSO₄·5H₂O) = 1 L · 0,3 mol/L.
n(CuSO₄·5H₂O) = 0,3 mol.
m(CuSO₄·5H₂O) = n(CuSO₄·5H₂O) · M(CuSO₄·5H₂O).
m(CuSO₄·5H₂O) = 0,3 mol · 249,7 g/mol.
m(CuSO₄·5H₂O) = 74,9 g.
Lithium
Li → Li⁺ + e⁻
Li 1s²2s¹ → Li⁺ 1s² + e⁻