Answered

Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

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

Sagot :

Answer:

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 .

Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.