A mixture formed by adding 40.0 mL of 2.5×10−2M HCL to 170mL of 1.0×10−2 HI. What would the pH be?

HCl completely dissociates in aqueous solution to give H₃O⁺ ions.
Number of moles of H₃O⁺ ions contributed by HCl = (2.5 × 10⁻² mol/L) × (40.0/1000 L) = 0.001 mol

HI also completely dissociates in aqueous solution to give H₃O⁺ ions.
Number of moles of H₃O⁺ ions contributed by HI = (1.0 × 10⁻² mol/L) × (170/1000 L) = 0.0017 mol

Total number of moles of H₃O⁺ ions = (0.001 + 0.0017) mol = 0.0027 mol
Total volume of the final solution = (40.0 + 170) mL = 210 mL = 0.21 L
[H₃O⁺] in the final solution = (0.0027 mol) / (0.21 L) = 0.0027/0.21 M

pH = -log[H₃O⁺] = -log(0.0027/0.21) = 1.89