Find the pH of a solution when 0.01 M HCl and 0.1 M NaOH are mixed in equal volumes

The correct option is 12.65.

Let's understand the step-by-step process to find the pH of the mixture.

Step 1: Understand the reaction between the acid and the base
When hydrochloric acid (HCl), a strong acid, and sodium hydroxide (NaOH), a strong base, are mixed, they neutralize each other according to the following chemical reaction:
HCl + NaOH ⇒ NaCl + H₂O

Step 2: Calculate the millimoles (or moles) of H⁺ and OH^- ions
Let the volume of each solution mixed be V mL (since they are mixed in equal volumes).
Molarity of HCl = 0.01 M
Molarity of NaOH = 0.1 M

Millimoles of H⁺ ions from HCl = Molarity × Volume (in mL) = 0.01 × V = 0.01V mmol
Millimoles of OH^- ions from NaOH = Molarity × Volume (in mL) = 0.1 × V = 0.1V mmol

Step 3: Determine the excess reactant
Since OH^- ions (0.1V mmol) are in excess compared to H⁺ ions (0.01V mmol), all the H⁺ ions will be neutralized, leaving excess OH^- ions in the solution.

Remaining millimoles of excess OH^- ions = Millimoles of OH^- - Millimoles of H⁺
Remaining millimoles of OH^- = 0.1V - 0.01V = 0.09V mmol

Step 4: Calculate the concentration of OH^- ions in the mixture
The total volume of the mixture is the sum of the volumes of both solutions:
Total Volume = V + V = 2V mL

The molar concentration of excess OH^- ions is:
$$[{\text{OH}}^{-}]=\frac{\text{Remaining millimoles of}{\text{OH}}^{-}}{\text{Total Volume in mL}}$$
$$[{\text{OH}}^{-}]=\frac{0.09V}{2V}=0.045\text{M}$$

Step 5: Calculate the pOH of the solution
The pOH is defined as the negative logarithm of the hydroxide ion concentration:
pOH = -log[OH^-]
pOH = -log(0.045)
pOH = -log(4.5 × 10^-2) = 2 - log(4.5)
Since log(4.5) is approximately 0.653:
pOH ≈ 2 - 0.653 = 1.35

Step 6: Calculate the pH of the solution
Using the relationship between pH and pOH at 25 °C:
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 1.35 = 12.65

Thus, the pH of the resulting solution is 12.65.