The pH of a solution obtained by mixing 100 cm³ of 0.1 M HCl solution with 100 cm³ of 0.2 M NaOH solution is

Step 1: Moles of HCl
Moles = concentration × volume
$Moles of HCl = 0.1~mol/dm^3 \times 0.1~dm^3 = 0.01~mol$

Step 2: Moles of NaOH
Moles = concentration × volume
$Moles of NaOH = 0.2~mol/dm^3 \times 0.1~dm^3 = 0.02~mol$

Step 3: Excess OH⁻
NaOH is in excess, so excess moles of OH⁻ = moles NaOH - moles HCl
$0.02~mol - 0.01~mol = 0.01~mol$

Step 4: Total volume of solution
$100~cm^3 + 100~cm^3 = 200~cm^3 = 0.2~dm^3$

Step 5: Concentration of OH⁻
$[OH^-] = \frac{0.01~mol}{0.2~dm^3} = 0.05~mol/dm^3$

Step 6: pOH
$pOH = -\log_{10} (0.05)$
$0.05 = 5 \times 10^{-2}$
$\log_{10} (5 \times 10^{-2}) = \log_{10} 5 + \log_{10} (10^{-2}) = 0.6990 - 2 = -1.3010$
$pOH = -(-1.3010) = 1.3010 \approx 1.3$

Step 7: pH
At 25°C, $pH + pOH = 14$
$pH = 14 - 1.3 = 12.7$

Final answer: 12.7