As shown by Henderson-Hasselbach equation, when concentrations of the conjugate base and the un-dissociated acid are equal, the pH of the solution equals pK a of the buffer pH = pK a + log 1 log 1 = 0, Therefore, the pH of the buffer solution is 7.38. A buffer is selected on the basis of its pK a and its chemical nature. The equation can be derived from the formula of pK a for a weak acid or buffer. The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in an acid-base reaction. Example 2. We have the equation, pH = pKa + log[Salt] / [Acid] log[Salt] / [Acid] = pH – pKa = 5 – 4.74 Henderson-Hasselbalch. We can also use the alkaline buffer equation to calculate the pH but need to take note on the following points: - pK b is for base CH 3 COO-which we can determine from pK a of CH 3 COOH via the formula . For maximum buffer capacity. Solution. Calculate the volume of 0.2M solution of acetic acid that needs to be added to 100 ml of 0.2M solution of sodium acetate to obtain a buffer solution of pH 5.00. pKa of acetic acid is 4.74. If we substitute the values in equation 1 above, we will get: pH = -log (4.2 x 10 -7 )+ log (0.035/0.0035) pH = 6.38 + 1 = 7.38. pH = pKa = -log Ka = -log (1.8 × 10-4) = 3.74. buffer capacity = 3.74. For solutions of a weak bases sometimes it is more convenient to use equation in the form. pK a + pK b = 14 - base is CH 3 COO-hence "salt" must be CH 3 COOH which is the conjugate acid partner of CH 3 COO-Interestingly the pH calculated using the alkaline buffer equation will give the … It can be used for pH calculation of a solution containing pair of acid and conjugate base - like HA/A -, HA - /A 2- or B + /BOH. 15.3. This is so called Henderson-Hasselbalch equation (or a buffer equation ). The balanced equation for an acid dissociation is: HA ⇌ H+ +A− HA ⇌ H + + A − This answer is the same one we got using the acid dissociation constant expression.