ma quiz _5 _ Titration_Ksp Practice KEYY

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

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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 ---

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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?
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