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For 100.0 mL of a solution that is 0.040M CH3COOH and 0.010 M CH3COO, what would be the pH after adding 10.0 mL 50.0 mM HCl?

Sagot :

Jufsanabriasaidn

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 is 3.90

Mickymike92idn

The pH of the buffer after addition of 10.0 mL of 50.0mM HCl is 3.90

What is the Henderson-Hasselbach equation?

The Henderson-Hasselbach equation is given as:

  • pH = pKa + log [A-] / [HA]

where

  • pKa of acetic acid (4.75),
  • [A-] is moles of the conjugate base
  • [HA] the moles of thw weak acid.

How to determine the moles of CH3COO- and CH3COOH

The formula for calculating number of moles is:

  • Moles = concentration × volume

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

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