What is the ph of a solution after the addition of 30.0 ml of 0.100 m naoh to 50.0 ml of 0.10 m hbr?

pH = 1.6

Explanation:

0 mL NaOH:

⇒ [ H3O+] = M HBr = 0.1 M

⇒pH = -log [H3O+] = 1

30 mL NaOH:

⇒ mol NaOH = 0.1 mol / L * 0.03 L =  3 E-3 mol

⇒ mol HBr = 0.05 L * 0.1 mol/L = 5 E-3 mol

⇒ M HBr = ( 5 E-3 mol - 3 E-3 mol) / 0.08 L = 0.025 M

⇒ pH = - log (0.025) = 1.6

2b2t

explanation:

2b2t