Answer:
The pH of the buffer is 3.90
Explanation:
The mixture of a weak acid CH3COOH and its conjugate base CH3COO produce a buffer that follows the equation:
pH = pKa + log [A-] / [HA]
Where pH is the pH of the buffer, pKa is the pKa of acetic acid (4.75), and [A-] could be taken as the moles of the conjugate base and [HA] the moles of thw weak acid.
To solve this question we need to find the moles of the CH3COOH and CH3COO- after the reaction with HCl:
CH3COO- + HCl → CH3COOH + Cl-
The moles of CH3COO- are its initial moles - the moles of HCl added
And moles of CH3COOH are its initial moles + moles HCl added
Moles CH3COO-:
Initial moles = 0.100L * (0.010mol / L) = 0.00100moles
Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles
Moles CH3COO- = 0.000500 moles
Moles CH3COOH:
Initial moles = 0.100L * (0.040mol / L) = 0.00400moles
Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles
Moles CH3COO- = 0.003500 moles
pH is:
pH = 4.75 + log [0.000500] / [0.00350]
pH = 3.90
The pH of the buffer after addition of 10.0 mL of 50.0mM HCl is 3.90
The Henderson-Hasselbach equation is given as:
where
The formula for calculating number of moles is:
The equation of the reaction is given below:
CH3COO- + HCl → CH3COOH + Cl-
moles of CH3COO- = initial moles - moles of HCl added
moles of CH3COOH = initial moles + moles HCl added
Moles CH3COO-
Molarity = 0.010 M
volume = 100 mL = 0.100 L
Initial moles = 0.100 L * 0.010 M = 0.001 moles
Moles HCl = 0.010L * 0.050 M= 0.0005 moles
Moles CH3COO- = 0.001 - 0.0005 moles
Moles of CH3COO- = 0.000500 moles
Moles CH3COOH:
Molarity = 0.040 M
volume = 100 mL = 0.100 L
Initial moles = 0.100L * (0.040mol / L) = 0.00400moles
Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles
Moles CH3COO- = 0.004 - 0.0005 moles
Moles CH3COO- = 0.003500 moles
Substituting the calculated values:
pH = 4.75 + log [0.000500] / [0.00350]
pH = 3.90
Therefore, the pH of the buffer after addition of the 10.0 mL 50.0mmHg HCL is 3.90
Learn more about about buffers and pH at: https://brainly.com/question/11851669
