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1) if we add 0.50 mL of 0.087 M HCL to the solution in part A . How many moles of acetic acid , sodium acetate, and NaOH will be present after the reaction is done?Part A states : A buffer solutions contains .120 M acetic acid and 0.150 M sodium acetate. How many moles of acetic acid and of sodium acetate are present in 50.0 ml of solution ? = 0.0075 moles.

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Lelia Lubowitz
Lelia LubowitzLv2
28 Sep 2019
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Related questions

If 10.0 mL of 1.0 M NaOH is added to 1.0 L of each of the following solutions, which shows the greatest change in pH?

A) 0.10 M acetic acid and 0.10 M sodium acetate

B) 0.50 M acetic acid and 0.50 M sodium acetate

C) 0.50 M sodium acetate

D) 0.50 M acetic acid

E) 0.10 M acetic acid

Calculate the pH of 75mL of the undiluted buffer solution after the addition of 15mL of 3M HCl.

The undiluted buffer solution is prepared by mixing 100mL of 1.000M sodium acetate with 100 mL of1.0798M acetic acid.

I calculated the moles of acetic acid to be 1.0798M x (.1L/.2L) x 0.075L = 0.0405

moles sodium acetate: 1.000M (.1/.2) x 0.075L = 0.0375

moles HCL = 3.00 x 0.015 = 0.045 mols

However if i plug in the value PH=4.75 + log (-0.0075/0.0855), I get a math error for the log value.. What did i do wrong?

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