5. What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of an aqueous w…
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Question “5. What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of an aqueous w…”
5. What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of an aqueous weak acid requires 29.80 mL of 0.0567 M NaOH? Ka = 3.2 x 10-4 for the weak acid.
Answer
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1L 0.0567 mol No moles of NaOH added = 29.80mL = 0.0567M 25 mL 25 mL = 0.0676M 25 mL 25 mL 25 mL 25 mL 25 mL 25 mL 25 mL 25 mL 25.76×10-‘M K 1.00×10-11 Baseionization constant, [HX]OH-] Ko = (x) (x) (x) (x) (x10-14 = 1.02×10-7 = 1.02×10-1 pH = 0.01×10 mL 101 = 0.02×10-1 = 0.02×10-1 pH = 0.01×10 1L1L *1000 mol 10-1 pH = 0.0676×10-7 = 7.977 mL = 0.0676×10-1 = 0.0676×10-‘M = 0.0676×10-1 = 0.02×10-1 = 0.0676×10-7 = 7.99×10-11 = x10-1L 101 = 7.99 x10-1L = 0.02×10-1L = 0.02×10-1L = 976×10-1L = 0.02×10-1L = 0.02×10-1L = 11 x10-1 L 101 = 0 x10-1L = x10-1L = x10 x10-1L = 10-1L = x10-1L = x10-1L = 10-1L = 10-1L = x10-1L = x10-1L = 7.99 x10-1L = x10-1L = 10-1L = x10-1L = x10-1L = x10-1L = 10-1L = 10-1L101 10-1L 10-1L = 1L 10-1L = 10 x10-1L 10-1L101 10-1L = 10-1L101 2×10 10 1L = 1L = x8.02×10 x10 x10 10 10 10 10 x10 10 10 x10 10 1L 10 1L 0.0567 moles No moles added of NaOH = 29.80mL = 0.0676 M 25 mL Total volume = (25+29.8) L Concentration of X (x), [x] =0.0548 L Ice table x + H2O-HX+ OH-I(M) = 0.0305M + 3.2×10+ = 3.0×10-11 Baseionization coefficient, [HX]OH–] Ko = (x] 3.1×10-12 pH = 7.99×10-7 = 1.00×10-M K 1.00×10-14 pH = 1.02×10-7 pH = x10-7 pH = x10-7 pH = 1.02×10-7 pH = 10-log(1.02x10x10x10x10-1 = 7.
Conclusion
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