A 50.0-ml sample of 0.50 m hcl is titrated with 0.50 m naoh. what is the ph of the solution after 28.0 ml of naoh have been added to the acid?

The balanced reaction between base and acid is;

NaOH + HCl - - - > NaCl + H₂O

NaOH is a strong base and HCl is a strong acid therefore complete dissociation. Stoichiometry of acid to base is 1:1

the number of moles of base added - 0.5 M / 1000 mL/L x 28.0 mL = 0.014 mol

the number of acid moles present - 0.5 M / 1000 mL/L x 50.0 mL = 0.025 mol

acid reacts with base 1:1 ratio

therefore excess amount of acid present - 0.025 - 0.014 = 0.011 mol

total volume = 50.0 + 28.0 = 78.0 mL

[H⁺] = 0.011 mol/0.078 L = 0.14 M

pH = - log [H⁺]

pH = - log (0.14)

pH = 0.85