Calculate the volume of 1.00 M acetic acid solution and mass of sodium acetate needed to make 250 mL of a 0.450 M acetic acid/acetate buffer solution with a pH of 4.00. (The pKa is 4.74 for acetic acid).
I work these buffer problems with millimols instead of M.
You want 250 mL x 0.450M = 112.5 millimols of the buffer.
so equation 1 is base + acid = 112.5
Equation 2 comes from the HH equation.
pH = pKa + log (base)/(acid)
4.00 = 4.74 + log b/a
Solve for b/a and rearrange to
base = ?*a. This is equation 2.
Solve the two equations simultaneously for millimols base and acid, then convert mmols acid to mL of the 1.0 M and convert mmols base to g NaAc.
Dr bob can you please elaborate a little more I struggling to put it together
To find the volume of 1.00 M acetic acid solution needed, we can use the formula for dilution:
\(M_1 \times V_1 = M_2 \times V_2\)
Where:
\(M_1\) is the initial concentration of the solution (1.00 M)
\(V_1\) is the volume of the initial solution (unknown)
\(M_2\) is the final concentration of the solution (0.450 M)
\(V_2\) is the final volume of the solution (250 mL = 0.250 L)
Let's solve for \(V_1\):
\(1.00 \, \text{M} \times V_1 = 0.450 \, \text{M} \times 0.250 \, \text{L}\)
\(V_1 = \frac{0.450 \, \text{M} \times 0.250 \, \text{L}}{1.00 \, \text{M}}\)
\(V_1 = 0.1125 \, \text{L} = 112.5 \, \text{mL}\)
Therefore, 112.5 mL of the 1.00 M acetic acid solution is needed to make a 250 mL 0.450 M acetic acid/acetate buffer solution.
Now, let's move on to calculating the mass of sodium acetate needed. Sodium acetate (CH3COONa) dissociates in water to yield acetate ions (CH3COO-) and sodium ions (Na+). The balanced equation for the dissociation reaction is:
\(CH3COONa \rightarrow CH3COO- + Na+\)
Since we want the acetate concentration to be 0.450 M, and sodium acetate is the only source of acetate ions, the concentration of sodium acetate (CH3COONa) should also be 0.450 M.
To calculate the mass of sodium acetate, we can use the formula:
\(n = \frac{m}{M}\)
Where:
\(n\) is the number of moles of sodium acetate
\(m\) is the mass of sodium acetate (unknown)
\(M\) is the molar mass of sodium acetate
The molar mass of sodium acetate (CH3COONa) is calculated by adding up the atomic masses of each element:
\(M = 12.01 \, \text{g/mol} \times 2 + 1.01 \, \text{g/mol} \times 3 + 16.00 \, \text{g/mol} + 22.99 \, \text{g/mol}\)
\(M = 82.03 \, \text{g/mol}\)
Now, we can solve for the mass of sodium acetate (m):
\(0.450 \, \text{mol/L} \times 0.250 \, \text{L} \times 82.03 \, \text{g/mol} = m\)
\(m = 9.127 \, \text{g}\)
Therefore, approximately 9.127 grams of sodium acetate is needed to make 250 mL of a 0.450 M acetic acid/acetate buffer solution with a pH of 4.00.