A 10.0-mL sample of 0.200 M hydrocyanic acid (HCN) is titrated with 0.0998 M NaOH. What is the pH at the equivalence point? For hydrocyanic acid, pKa = 9.31.a) 7.00 b) 8.76 c) 9.31 d) 11.07
A 10.0-mL sample of 0.200 M hydrocyanic acid (HCN) is titrated with 0.0998 M NaOH. What is the pH at the equivalence point? For hydrocyanic acid, pKa = 9.31.a) 7.00 b) 8.76 c) 9.31 d) 11.07
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Chapter16: Acid-base Equilibria
Section: Chapter Questions
Problem 16.109QP: Find the pH of the solution obtained when 25 mL of 0.065 M benzylamine, C7H7NH2, is titrated to the...
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A 10.0-mL sample of 0.200 M hydrocyanic acid (HCN) is titrated with 0.0998 M NaOH. What is the pH at the equivalence point? For hydrocyanic acid, pKa = 9.31.
a) 7.00 b) 8.76 c) 9.31 d) 11.07
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Im confused as to how you got .002 moles for NaOH and CN-, I know there are .002 moles of HCN, but how did you get that for NaOH and CN-?
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