*should be quite close to the desired p H so that the ratio of base to acid in the Henderson-Hasselbalch equation will be close to 1.*As the ratio of base to acid deviates from 1, the addition of acids and bases to the buffer will have a more profound effect on the p H.

= 1.82 [A⁻] = 1.82[HA] Also, [A⁻] [HA] = 0.100 mol/L 1.82[HA] [HA] = 0.100 mol/L 2.82[HA] = 0.100 mol/L [HA] = 0.0355 mol/L [A⁻] = (0.100 – 0.0355) mol/L = 0.0645 mol/L You need 0.0355 mol of acetic acid and 0.0645 mol of sodium acetate to prepare 1 L of the buffer.

More than one problem with solutions with additional info here.

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.

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

The process for finding the p H of the mixture after a strong base has been added is similar to the addition of a strong acid shown in the previous section.

Next, we substitute Equation \(\ref\) into Equation \(\ref\), which gives the concentration of HA as \[\mathrm\label\] Finally, substituting Equation \(\ref\) and \(\ref\) into Equation \(\ref\) and solving for p H gives a general equation for a buffer’s p H.

\[\mathrm\] If the initial concentrations of the weak acid, Henderson and Hasselbalch Lawrence Henderson (1878-1942) first developed a relationship between \([H_3O^ ]\), \([HA]\), and \([A^−]\) while studying the buffering of blood.

The above equation for K In order to calculate the p H of the buffer solution you need to know the amount of acid and the amount of the conjugate base combined to make the solution.

These amounts should be either in moles or in molarities. Example: A buffer solution was made by dissolving 10.0 grams of sodium acetate in 200.0 m L of 1.00 M acetic acid.

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