Computing pH of 1 L solution containing acetic acid and Sodium acetate after adding HCl.

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Dhamnekar Winod
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Computing pH of 1 L solution containing acetic acid and Sodium acetate after adding HCl.

Post by Dhamnekar Winod »

A 1.0 L solution contains 0.2 M CH3COOH and 0.2 M CH3COONa. What is the pH after 0.05 M of HCl has been added to the solution?

What the effect would be on adding the HCl to 1 L containing 0.2 mol of CH3COOH alone (not a buffer)?

In other words, What is the pH of 1 L solution containing 0.2 M of CH3COOH alone after adding 0.05 M HCl?

Please explain in details the computations.
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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ChenBeier
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Re: Computing pH of 1 L solution containing acetic acid and Sodium acetate after adding HCl.

Post by ChenBeier »

You didn't tell what is the volume of the 0,05 M HCl you add. So you can not calculate.
In a) you get a new buffer using Henderson Hasselbalch equation.
Pure acetic acid has no impact. At least only HCl gives the pH.
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Dhamnekar Winod
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Re: Computing pH of 1 L solution containing acetic acid and Sodium acetate after adding HCl.

Post by Dhamnekar Winod »

Hello,
Consider the volume of HCl is negligible. Would you show me your computations of pH in the both cases, if possible?
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: Computing pH of 1 L solution containing acetic acid and Sodium acetate after adding HCl.

Post by ChenBeier »

You have to calculate pH = pKa - log ((nCH3COOH + nHCl)/(nNaCH3COO - nHCl))
n = c* V , V you dont know, you cannot calculate.
If only acetic acid, hydrochloric acid and no Acetate is present then pH is driven by cHCl, pH = -log(cH+) = -log(cHCl)
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