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Sagot :
Answer:
pH = 4.71
Explanation:
We can find the pH of a buffer (Mixture of weak acid: CH3COOH, and its conjugate base: CH3COONa) using H-H equation:
pH = pKa + log [CH3COONa] / [CH3COOH]
Where pH is the pH of the buffere = 4.74, pKa the pka of the buffer and [] could be taken as the moles of each reactant.
As initially [CH3COONa] = [CH3COOH], [CH3COONa] / [CH3COOH] = 1:
pH = pKa + log 1
4.74 = pKa
To solve this question we need to find the initial moles of each species, The CH3COONa reacts with HCl to produce CH3COOH. That means the moles of CH3COOH after the reaction are: Initial CH3COOH + Moles HCl
Moles CH3COONa: Initial CH3COONa - Moles HCl.
Moles CH3COOH:
0.100L * (0.50mol / L) = 0.050 moles CH3COOH + 0.0020 moles HCl =
0.052 moles CH3COOH
Moles CH3COONa:
0.100L * (0.50mol / L) = 0.050 moles CH3COONa - 0.0020 moles HCl =
0.048 moles CH3COONa
Using H-H equation:
pH = 4.74 + log [0.048 moles] / [0.052 moles]
pH = 4.71
The pH be after 0.0020 mol of HCl has been added to 100.0 mL of the buffer is 4.71.
What is buffer solution?
Buffer solution is a mixture of weak acid and its conjugate base or vice versa.
We can calculate the pH of buffer solution by using Henderson - Hasselbalch Equation:
pH = pKa + log[CH₃COONa] / [CH₃COOH]
Initially concentration of CH₃COONa is equal to the concentration of CH₃COOH and equal becomes:
4.74 = pKa + log(1)
pKa = 4.74
Given moles of added HCl moles = 0.002 mole
Given molarity of each CH₃COOH & CH₃COONa = 0.50M
Given Volume = 100mL
We can calculate the moles by using the formula:
n = M × V
Moles of CH₃COOH & CH₃COONa = 0.100 × 0.50 = 0.050 moles
Moles of CH₃COOH = 0.050moles CH₃COOH + 0.0020moles HCl =
0.052moles CH₃COOH
Moles of CH₃COONa = 0.050moles CH3COONa - 0.0020moles HCl =
0.048moles CH₃COONa
Now, resultant pH will be:
pH = 4.74 + log [0.048 moles] / [0.052 moles]
pH = 4.71
Hence, pH of resultant solution is 4.71.
To know more about Henderson - Hasselbalch Equation, visit the below link:
https://brainly.com/question/26746644
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