the pH of 10.00 mL of 0.100 M HCl after adding 10.00 mL of indicator solution

Answer 1

The 0.100 M HCl has been diluted from 10 mL to 20 mL; therefore,
(HCl) = 0.100 x (10/20) = 0.05M
pH = -log(H^+)
pH = -log(0.0500)
pH = -(-1.301) = 1.301