Consider nicotinic acid, HC6H4NO2 (HNic Ka = 1.4x10^-5) and its conjugate base,

Question

Consider nicotinic acid, HC6H4NO2 (HNic Ka = 1.4x10^-5) and its conjugate base, C6H4NO2^-1 (Nic^-1 Kb = 7.1x10^-10)

1. A buffer is prepared by dissolving 47.13 g of NaC6H4NO2 (NaNic) in 1.25 L of a solution of 0.295M HC6H4NO2 (HNic). Calculate the pH of the buffer.

2. 40.0 mL of .575M HNic is titrated with .335 M KOH.

A. Caluclate the pH of the solution before titrating.

B. What is the pH of the solution halfway to the equivalence point

C. Calculate the pH after .050L of KOH has been added.

D. Calculate the volume of KOH required to reach the equivalence point

E. Calculate the pH of the solution at the equivalence point.

Thank you so much for your help!

Explanation / Answer

Question 1.

This is a buffer; buffers are used to maintain pH when acid/base is added.

In solution we have:

HC6H4NO2(aq) <-> H+(aq) + C6H4NO2-(aq)

If we add any conjugate extra...

C6H4NO2-(aq) + H2O(l) <--> C6H4NO2(aq) + OH-(aq)

and H+ and OH- will neutralize to form water... this way we have a dynamic equilibrium

The generic formula for a buffer (Henderson Hasselbalch Equation)

pH = pKa + log([conjguate]/[acid])

Recall that

pKa = -log(Ka)

pKa = -log(1.4*10^-5) = 4.853

then we need to find all concentrations...

[conjugate] = mol of conjugate/ total volume

[acid] = mol of acid/ total volume (actually it is already given, no need to calculate)

Total V = 1.25 L

mol of conjugate = mass of conjugate / Molar Mass

MW of NaC6H4NO2 = 145.0912 g/mol

mol = 47.13/145.0912 = 0.32483 mol of conjugate

[conjugate] = mol/V = 0.32483/1.25 = 0.259864 M of conjugate

[acid] = 0.295 M of acid

now... substitute in pH equation

pH = pKa + log([conjguate]/[CID])

pH = 4.853 + log(0.259864/0.295) = 4.797

pH of solution after addition of salt = 4.797

For Question 2. Please consider posting multiple questions in multiple set of Q&A. We are not allowed to answer to multiple questions in a single set.

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