What is the pH of a solution prepared by mixing 150 mL of 0.1M HCl with 300mL of 0.1M sodium acetate (NaOAC) and diluting the mixture of 1L? PKa of acetic acid is 4.76. and give solution
4.77
4.77
To calculate the pH of the solution, we need to consider the acid-base properties of the HCl and sodium acetate solutions.
First, let's find the moles of HCl and sodium acetate in each solution.
Moles of HCl:
Volume of HCl solution = 150 mL = 0.15 L
Concentration of HCl = 0.1 M
Moles of HCl = Volume x Concentration = 0.15 L x 0.1 M = 0.015 moles
Moles of sodium acetate:
Volume of sodium acetate solution = 300 mL = 0.3 L
Concentration of sodium acetate = 0.1 M
Moles of sodium acetate = Volume x Concentration = 0.3 L x 0.1 M = 0.03 moles
Now, let's balance the equation representing the ionization of acetic acid:
CH3COOH ⇌ CH3COO- + H+
For each mole of acetic acid that ionizes, it forms one mole of acetate ion (CH3COO-) and one mole of hydrogen ion (H+).
Since the sodium acetate dissociates completely in water, we have 0.03 moles of acetate ions.
Next, let's calculate the concentrations of H+ and OH- ions using the concept of acid-base equilibrium.
The pKa of acetic acid is given as 4.76, which means the equilibrium constant (Ka) is 10^(-pKa) = 10^(-4.76).
Using the equation for Ka:
Ka = [H+][CH3COO-] / [CH3COOH]
We can simplify this to:
[H+][CH3COO-] = Ka x [CH3COOH]
After substituting the values, we get:
[H+][0.03] = 10^(-4.76) x [0.015]
[H+] = (10^(-4.76) x 0.015) / 0.03
Now, let's calculate the pH using the concentration of H+ ions.
pH = -log10[H+]
Substituting the value of [H+], we get:
pH = -log10[(10^(-4.76) x 0.015) / 0.03]
By solving this equation, you can find the pH of the solution prepared by mixing 150 mL of 0.1M HCl with 300 mL of 0.1M sodium acetate and diluting the mixture to 1L.
HCl = 150 x 0.1 = 15 millimols
NaOAc = 300 x 0.1 = 30 millimols.
pH = pKa + log (base)/(acid)
Substitute and solve for pH