Find the pH of solution prepared by mixing 25 ml of a 0.5 M solution of HCl, 10 ml of a 0.5 M solution of NaOH and 15 ml of water

We know that  HCl and NaOH reaction  is neutralization,( m.e. = milli-equivalent)∴   m.e.  of  HCl = 0.5 × 25=12.5 m.e. of NaOH = 0.5 × 10=5.0m.e. of HCl in the resultant mixture= 12.5 - 5.0 = 7.5Total volume = (25 + 10 + 15) ml = 50 ml∴  Normality of HCl = m.e.Vol(ml)=7.5050 ∴   Molarity = 7.5050 ∴    H+ = 7.5050 ∴    pH = -log  7.5050 = 0.8239