Question

A laboratory made a buffer solution by mixing 25 mL of 0.10 M acetic acid (Ka = 1.8 x 10^-5) with 15 mL of 0.20 M sodium acetate solution. Calculate the pH of the buffer solution assuming both the acid and the acetate ions are completely dissociated.

Answer 1

To calculate the pH of the buffer solution, we can use the Henderson-Hasselbalch equation:pH = pKa + log [A-]/[HA] where pKa is the negative logarithm of the acid dissociation constant  Ka , [A-] is the concentration of the conjugate base  acetate ion , and [HA] is the concentration of the weak acid  acetic acid .First, let's find the pKa:pKa = -log Ka  = -log 1.8 x 10^-5   4.74Next, we need to find the final concentrations of acetic acid and acetate ion in the buffer solution. To do this, we can use the formula:M1V1 = M2V2For acetic acid: 0.10 M  25 mL  = M2 40 mL M2 =  0.10 M  25 mL  / 40 mL  0.0625 MFor sodium acetate: 0.20 M  15 mL  = M2 40 mL M2 =  0.20 M  15 mL  / 40 mL  0.075 MNow we can plug these values into the Henderson-Hasselbalch equation:pH = 4.74 + log 0.075 / 0.0625   4.74 + 0.17  4.91So, the pH of the buffer solution is approximately 4.91.