Answer:
[tex]\boxed{5.01\times 10^{-9}\text{ mol/L}}[/tex]
Explanation:
pOH = 5.30
pH = 14.00 - 5.30 = 8.70
pH = -log[H₃O⁺]
[tex]\begin{array}{rcl}[\text{H}_{3}\text{O}^{+}] & = & 10^{- \text{pH}} \text{ mol/L}\\ & = & 10^{-8.30}\text{ mol/L}\\ & = &\boxed{\mathbf{5.01\times 10^{-9}}\textbf{ mol/L}}\\\end{array}[/tex]
