If 50.0 mL of a 0.75 M hydrochloric acid solution is titrated with 1.0 M sodium hydroxide, what is the pH after 10.0 mL of NaOH have been added

Explanation:

moles of hydrochloric acid = 50 x 10⁻³ x .75 = 37.5 x 10⁻³ moles in HCl solutions.

moles of NaOH in its solution = 10 x 10⁻³ x 1 moles = 10 x 10⁻³ moles .

10 x 10⁻³ moles  of NaOH will neutralise 10 x 10⁻³ moles of HCl .

HCl remaining = (37.5 - 10 ) x 10⁻³ moles = 27.5 x 10⁻³ moles

volume of solution = 50 + 10 = 60 mL = 60 x 10⁻³ L

molarity of remaining HCl = 27.5 x 10⁻³ / 60 x 10⁻³

= .4583 M .HCl

concentration of H⁺ = .4583 M

pH = - log ( .4583 )

= .34 .

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