A buffer solution is prepared from 1.00 L of 0.0500 M CH3COOH (Ka = 1.8 10–5) and 2.50 g sodium ac — Acid-Base Equilibrium / Buffers Chemistry Question
Problem Context
A buffer solution is prepared from 1.00 L of 0.0500 M CH3COOH (Ka = 1.8 10–5) and 2.50 g sodium acetate, Na(CH3COO).
What is the pH of this solution?
Model Answer
1.00 L 0.050 mol L-1 = 0.0500 mol CH3COOH
2.50 g Na[CH3COO]/82.04 g mol-1 = 0.0305 mol CH3COO–
pH = pKa + log([CH3COO– ]/[CH3COOH]) = 4.74 + log(0.0305/0.0500) = 4.53
1.00 mL of a 1.00 M solution of hydrochloric acid is added to the buffer. Write an equation for the major reaction that takes place, and calculate the pH after addition.
Model Answer
CH3COO– (aq) + H3O+ (aq) CH3COOH(aq) + H2O(l)
pH = 4.74 + log({0.0305 – 0.001}/{0.0500 + 0.001}) = 4.50
Calculate the mass of sodium hydroxide that would need to be added to 1.00 L of 0.050 M CH3COOH to produce a solution with the same pH as the original buffer.
Model Answer
To achieve the same pH, one would need the same ratio of conjugate acid to conjugate base, so (mol CH3COOH)/(mol CH3COO– ) = 1.64.
If x is the number of mol NaOH added, then (0.0500 mol – x)/x = 1.64
0.0500 mol = 2.64x
x = 0.0189 mol NaOH
0.0189 mol NaOH 40.0 g mol-1 = 0.756 g NaOH
Suppose the buffer had been made with 1.00 L of 0.0500 M chloroacetic acid, ClCH2COOH, and 2.50 g sodium chloroacetate, Na(ClCH2COO). Compared to the original buffer, would this buffer have a lower pH, a higher pH, or the same pH, or can one not draw a qualitative conclusion without quantitative information about the Ka of ClCH2COOH? Justify your answer.
Model Answer
The electron-withdrawing character of Cl compared with H means that ClCH2COOH will be more acidic (have a lower pKa) than CH3COOH. The higher molar mass of Na(ClCH2COO) means that one is adding fewer moles of base to the chloroacetic acid buffer than one did to the acetic acid buffer (the number of moles of acid are the same in both cases). Both these factors will contribute to a lower pH of the ClCH2COOH/ClCH2COO– buffer.