Question

Equal volumes of two solutions, one containing a strong acid at pH 2 and the other containing a strong base at pH 12, are mixed. Once equilibrium is achieved, what is the final pH of the combined solution?

Answer 1

Answer:

7  

Explanation:

Assume we have 1 L of each solution.

Solution 1

$\text{[H ^{+} ]}= 10^\text{-pH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of H}^{+} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol H}^{+}}{\text{1 L solution}} = 10^{-2}\text{ mol H}^{+}$

Solution 2

pH = 12

pOH = 14.00 - pOH = 14.00 - 12 = 2.0

$\text{[OH ^{-} ]}= 10^\text{-pOH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of OH}^{-} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol OH}^{-}}{\text{1 L solution}} = 10^{-2}\text{ mol OH}^{-}$

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