Answer:
7
Explanation:
Assume we have 1 L of each solution.
Solution 1
[tex]\text{[H$^{+}$]}= 10^\text{-pH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of H}^{+} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol H}^{+}}{\text{1 L solution}} = 10^{-2}\text{ mol H}^{+}[/tex]
Solution 2
pH = 12
pOH = 14.00 - pOH = 14.00 - 12 = 2.0
[tex]\text{[OH$^{-}$]}= 10^\text{-pOH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of OH}^{-} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol OH}^{-}}{\text{1 L solution}} = 10^{-2}\text{ mol OH}^{-}[/tex]
3. pH after mixing
H⁺ + OH⁻ ⟶ H₂O
I/mol: 10⁻² 10⁻²
C/mol: -10⁻² -10⁻²
E/mol: 0 0
The H⁺ and OH⁻ have neutralized each other. The pH will be that of pure water.
pH = 7
