Solving Buffer Problems

Solving Buffer Problems-12
This prevents the p H of the solution from significantly rising, which it would if the buffer system was not present.

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Kurt Hasselbalch (1874-1962) modified Henderson’s equation by transforming it to the logarithmic form shown in Equation \(\ref\).

The assumptions leading to Equation \(\ref\) produce a minimal error in p H ( are \[C_\mathrm=\dfrac\] \[\mathrm\] \[C_\mathrm=\dfrac\] \[\mathrm\] Substituting these concentrations into the Equation \(\ref\) gives a p H of \[\textrm=9.24 \log\dfrac=9.10\] With a p H of 9.06, the concentration of HCl, the approximations leading to Equation \(\ref\) are reasonable.

A more useful relationship relates a buffer’s p H to the initial concentrations of the weak acid and the weak base.

We can derive a general buffer equation by considering the following reactions for a weak acid, HA, and the salt of its conjugate weak base, Na A.

Example: 50.0 m L of 0.100 M HCl was added to a buffer consisting of 0.025 moles of sodium acetate and 0.030 moles of acetic acid.

Solving Buffer Problems

What is the p H of the buffer after the addition of the acid?

Calculations are based on the equation for the ionization of the weak acid in water forming the hydronium ion and the conjugate base of the acid.

"HA" represents any weak acid and "A [HA] A buffer system can be made by mixing a soluble compound that contains the conjugate base with a solution of the acid such as sodium acetate with acetic acid or ammonia with ammonium chloride.

The maximum amount of base that can be added is equal to the amount of weak acid present in the buffer.

Example: What is the maximum amount of acid that can be added to a buffer made by the mixing of 0.35 moles of sodium hydrogen carbonate with 0.50 moles of sodium carbonate?

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