What is the pH at the equivalence point in the titration of a 27.9 mL sample...

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science chemistry

Answer 1

Answer #1

HCN + NaOH ------------> NaCN + H2O

millimoles of acid = 27.9 x 0.301 = 8.40

8.40 millimoles NaOH must be added to reach equivalence point

8.40 = V x 0.358

V = 23.5 mL

total volume = 23.5 + 27.9 = 51.4 mL

[NaCN] = 8.40 / 51.4 = 0.163 M

for salt of weak acid

pH = 1/2 [pKw + pKa + log C]

pKa of HCN = 9.2

pH = 1/2 [14 + 9.2 + log 0.163]

pH = 11.2

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Answer 2

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