ma quiz _5 _ Titration_Ksp Practice KEYY

ma quiz _5 _ Titration_Ksp Practice KEYY - Titration...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Titration Practice 100 ml of HClO solution with an unknown concentration is titrated with 0.0123 M LiOH. The titration requires 50 ml of LiOH to reach the equivalence point. (HClO k a = 3.0x10 -8 ) a. What is the concentration of the unknown HClO solution? (100ml) (x) = (0.0123M) (50ml) x = .615M/100 [HClO] = .00615M b. What is the pH of the acid before any base has been added? HClO + H 2 O ClO - + H + I .00615M --- 0 0 C -x --- +x +x E .00615M-x --- x x 3.0x10 -8 = x 2 /(.00615 – x) 1.845x10 -10 – 3.0x10 -8 x = x 2 x 2 + 3.0x10 -8 -1.845x10 -10 = 0 x or [H + ] = 1.3568x10 -5 pH = 4.87 c. What is the pH when 20 ml of base has been added? .02 x (.0123M) = 2.46x10 -4 moles LiOH .1 x (.00615M) = 6.15x10 -4 moles HClO HClO + LiOH LiClO + H 2 O B 6.15x10 -4 0 0 --- C (6.15x10 -4 )-( 2.46x10 -4 ) 2.46x10 -4 0 --- A 3.69x10 -4 0 2.46x10 -4 ---
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
pH = pKa + log (base/acid) pH = -log(3.0x10 -8 ) + log (2.46x10 -4 /3.69x10 -4 ) pH = 7.5229 + log (.66667) pH = 7.35 d. What is the pH at equivalence point? HClO + LiOH LiClO + H 2 O B 6.15x10 -4 0 0 --- C 0 6.15x10 -14 0 --- A 0 0 6.15x10 -14 --- 6.15x10 -14 moles/.15L = 4.1x10 -3 M LiCl K w = K a x K b k b = 1.0x10 -14 /3.0x10 -8 k b = 3.33x10 -7 LiClO + H 2 0 HClO + LiOH I 4.1x10 -3 --- 0 --- C -x --- +x +x E 4.1x10 -3 –x --- x x 3.33x10 -7 = x 2 /4.1x10 -3 –x x 2 + 3.33x10 -7 x – 1.3653x10 -9 = 0 x or [OH] = 3.67834x10 -5 pOH = 4.43 pH = 14 – 4.43 pH = 9.57 e. What is the pH of the solution when 51 ml of base has been added?
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern