How many moles of sodium acetate CH3COONA must be dissolved in 1.00L of 0.200M acetic acid to make a buffer pH 4.56? ka of acetic acid is 1.76x10^-
Use the Henderson-Hasselbalch equation.
PH= 4.56, PKa= 4.74
PH = pka +log(Base/Acid)
-0.18 = log(Base/0.200)
B = 0.1321 M >>>> o.1321 mol/L
Molocular weight of acetic acid = 82.03 g
(0.1321 mol/L)(0.4 l) 0.05284 mol
(0.05284 mol)(82.03 g/ mol) = 4.33 g.
I hope this helps!
To determine the number of moles of sodium acetate (CH3COONA) needed to make a buffer solution, we need to calculate the concentration of CH3COONA required.
The Henderson-Hasselbalch equation can be used to solve this problem, which is given by:
pH = pKa + log([A-]/[HA])
Where:
- pH is the desired pH of the buffer solution (4.56)
- pKa is the negative logarithm of the acid dissociation constant (ka) of acetic acid (1.76x10^-5)
- [A-] is the concentration of the conjugate base (CH3COO-) of acetic acid
- [HA] is the concentration of the acid (acetic acid)
Rearranging the Henderson-Hasselbalch equation, we can solve for [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)
Now, let's calculate the value of [A-]/[HA]:
[A-]/[HA] = 10^(4.56 - (-log10(1.76x10^-5)))
[A-]/[HA] = 10^(4.56 + 4.7549)
[A-]/[HA] = 10^9.3149
Next, we need to consider the stoichiometry of the balanced equation for the dissociation of acetic acid:
CH3COOH ⇌ CH3COO- + H+
One mole of acetic acid (CH3COOH) reacts to form one mole of the conjugate base (CH3COO-). Therefore, the stoichiometric ratio of [A-] to [HA] is 1:1.
Given that [A-]/[HA] = 10^9.3149, we can conclude that the concentration of [A-] is the same as [HA]. Therefore, the concentration of sodium acetate (CH3COONA) needed is given by:
[A-] = [HA] = 10^9.3149
Finally, to calculate the number of moles of CH3COONA required, we multiply the concentration with the volume of the solution:
Number of moles = Concentration × Volume
Number of moles = 10^9.3149 × 1.00 L
Now you can calculate the actual numerical value using a calculator.