e) If the solution in part a) is titrated with 0.250 M NaoH, what volu equired t

Question

e) If the solution in part a) is titrated with 0.250 M NaoH, what volu equired to reach the equivalence point, i.e, to neutralized all of the benzoic acid? a) is titrated with 0.250 M NaOH, what volume, in L, of NaOH is total volume of the combined benzoic acid and NaOH solutions at the equivalence point? 8) What is the concentration of benzoate, [A], at the equivalence point? h) From the K. given in part b), calculate the K, of benzoate [A']. i) What is the [OH] at the equivalence point? A) H20 ?? ?? j) What is the pH at the equivalence point? k) From the pK, given in part b), what ratio of benzoate to benzoic acid is needed t buffer of pH -4.00? o prepare a 9) The Kap of BaF2 is 1.8 X 10, What is the molar solubility?

Explanation / Answer

Part E

Moles of benzoic acid = 0.01 mol

Molarity of NaOH = 0.250 M

The balanced reaction

C6H5COOH + NaOH = C6H5COONa + H2O

Molar ratio of C6H5COOH : NaOH = 1 : 1

Moles of NaOH required = 0.01 mol

Volume of NaOH required = moles/molarity

= 0.01 mol / (0.250 mol/L)

= 0.0400 L

Part f

Total volume of solution = volume of Benzoic acid + volume of NaOH

= 0.100 + 0.0400 = 0.140 L

Part g

Concentration of benzoate

= moles of benzoate formed / total volume

= 0.01 mol / 0.140 L

= 0.0714 M

Part h

Ka = 6.46 x 10^-5

Kw = Ka x Kb

10^-14 = 6.46 x 10^-5 x Kb

Kb = (10^-14) /(6.46 x 10^-5)

= 1.548 x 10^-10

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