Question

A compound has a pka of 7.4. to 100 ml of a 1.0 m solution of this compound at ph 8.0 is added 30 ml of 1.0 m hydrochloric acid. the resulting solution is ph:

Answer 1

 Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.

Answer 2

100ml of 1M compound with 7.4 pka has pH 8.0. 30 ml of 1.0 m hydrochloric acid is added to the solution. The resulting solution is pH will be 7.4.

Using the Henderson-Hasselbalch equation

$\bold { pH = pKa + log \dfrac { [A-]}{[HA]} }$

Before HCl addition

$\bold {8.0 = 7.4 + log \dfrac {[A-]}{[HA]} }$

[A-]/[HA] = 4.0...........1

Also, 0.1 L  = 0.1 moles (mole/L)

Therefore, [A-] + [HA] = 0.1 ..........2

Solving the equations 1 and 2,

A- = 0.08 moles

AH = 0.02 moles

Adding strong acid reduces

[A-] and increases [AH] by the same amount. 0.03 L  = 0.03 moles, when HCl added,

So,

A- = 0.08 - 0.03 = 0.05 moles

AH = 0.02 + 0.03 = 0.05 moles

Therefore, after HCl addition,

$\bold {\dfrac {[A^-]}{[HA]} = \dfrac {0.05 }{0.05} = 1.0 }$

Put the value in the equation, we get,

pH = 7.4 + log(1.0) = 7.4

Therefore the final pH of the solution is 7.4.

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