Given .300 L of a buffer solution that is .250 M HC2H3O2 and .50 M NaC2H302, what is the new pH if .0060 mol HCL is added? what about .0060 mol NaOH?
The Henderson-Hasselbalch equation works all buffer problems.
i see that the equation is pH=pKa+log[base]/[acid]. but i do not know how to get pKa
The acid is acetic acid. Either in your problem or in your text the Ka should be listed as close to 1.8E-5 which is a pKa of 4.74 (pKa = -log Ka). Use the value in your text or notes, not the one I used. The base is acetate. The acid is acetic acid in the HH equation.
To determine the new pH of the buffer solution after adding .0060 mol of HCl or NaOH, we need to consider the reaction that occurs between the added compound and the components of the buffer solution. In this case, we have a buffer solution composed of HC2H3O2 (acetic acid) and NaC2H302 (sodium acetate). Let's calculate the new pH step by step for each scenario:
1. Adding .0060 mol of HCl:
First, we need to determine which component of the buffer will react with the HCl. As HCl is a strong acid, it will completely dissociate in water, causing an increase in H+ concentration, and consequently, a decrease in pH. Acetic acid (HC2H3O2) will react with the added HCl, forming additional water and acetate ions. The balanced equation for this reaction is:
HC2H3O2 + HCl → H2O + C2H3O2-
Since .0060 mol of HCl is added, an equal amount of mol of H+ ions will be formed. The initial concentration of acetic acid is .250 M and assuming its concentration change is negligible, it remains the same after the reaction. The concentration of acetate ions, however, will change accordingly. The final concentration of acetate ions can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([acetate]/[acetic acid])
The pKa value of acetic acid is 4.74.
[acetate] = (moles of acetate ions) / (total volume of the solution)
[acetic acid] = initial concentration of acetic acid
2. Adding .0060 mol of NaOH:
NaOH is a strong base that will react with the acidic component of the buffer solution, which is acetic acid (HC2H3O2). This reaction will lead to the formation of water and acetate ions. The balanced equation is given by:
NaOH + HC2H3O2 → H2O + NaC2H3O2
By adding .0060 mol of NaOH, an equal amount of moles of OH- ions will be formed. The initial concentration of acetic acid is again .250 M and assuming its concentration change is negligible, it remains the same after the reaction. The concentration of acetate ions will change accordingly. Using the Henderson-Hasselbalch equation mentioned above, the new pH can be calculated.
Please note that if the added substance is a strong acid or base, the buffer capacity may be overwhelmed, and the pH will significantly shift towards the pH of the strong acid or base added. In such cases, the Henderson-Hasselbalch equation might not accurately represent the pH change.