Calculate pH of 0.01 M CH3COONa solution if pKa of CH3COOH is 4.74.
Ph= 7+1/2(pKa+logC)
= 7+1/2(4.7+log10^-2)
= 7+2.35-1
=8.35
To calculate the pH of a solution of sodium acetate (CH3COONa), we need to take into account the dissociation of sodium acetate into its respective ions:
CH3COONa → CH3COO- + Na+
Sodium acetate is the salt of acetic acid (CH3COOH) and sodium hydroxide (NaOH). In water, CH3COO- acts as a base and accepts proton (H+) from water to form acetic acid (CH3COOH). This results in the formation of hydroxide ions (OH-) in the solution.
Since the pKa of acetic acid (CH3COOH) is given as 4.74, we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log([A-]/[HA])
Where [A-] is the concentration of the conjugate base (CH3COO-) and [HA] is the concentration of the acid (CH3COOH).
In this case, since we have a 0.01 M solution of sodium acetate, we will assume that all of the sodium acetate dissociates, giving us a concentration of 0.01 M for both CH3COO- and CH3COOH. Therefore, the equation becomes:
pH = 4.74 + log(0.01/0.01)
Simplifying further:
pH = 4.74 + log(1)
Since the log of 1 is 0, the equation further simplifies to:
pH = 4.74 + 0
Therefore, the pH of the 0.01 M CH3COONa solution is equal to 4.74.
To calculate the pH of a solution containing a weak acid and its conjugate base, you can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Where:
pH is the acidity or basicity of the solution
pKa is the negative logarithm of the acid dissociation constant (Ka) of the weak acid
[A-] is the concentration of the conjugate base (CH3COO-) in moles per liter (M)
[HA] is the concentration of the weak acid (CH3COOH) in moles per liter (M)
In this case, you are given the pKa of CH3COOH (4.74) and the concentration of CH3COONa (0.01 M).
First, let's determine the concentrations of the weak acid and its conjugate base. Since CH3COONa is a strong electrolyte, it fully dissociates into its constituent ions in solution. CH3COONa dissociates into CH3COO- (conjugate base) and Na+ (sodium ion).
Therefore, [A-] = [CH3COO-] = 0.01 M (concentration of CH3COONa)
And [HA] = [CH3COOH] = 0 M (since the concentration of CH3COOH is not given)
Now, substitute these values into the Henderson-Hasselbalch equation:
pH = 4.74 + log (0.01/0)
Since the concentration of CH3COOH is 0 (not given), the logarithm of 0 is undefined. Therefore, the pH of the solution cannot be calculated with the given information.