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If 200 mL of a 0 .031 molar solution of H 2 SO 4 are added to 84 mL of a 0 .150 M KOH solution, what is the pH of the resulting solution?

Options

  1. A12.4
  2. B1.7
  3. C2.2
  4. D10.85

Correct answer

D. 10.85

Step-by-step solution

m m o l o f H + i n i t i a l = 200 × 0.031 × 2 = 12.4 m m o l o f O H - i n i t i a l = 84 × 0.15 = 12.6 m m o l o f O H - l e f t a f t e r n e u t r a l i s a t i o n = 0.2 OH − final   =   0 .2 284   =   7   ×   10 − 4   M pOH = 3 .15 and pH = 10 .85

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