February 22, 2017

Homework Help: chemistry

Posted by Joseph on Wednesday, April 9, 2008 at 11:03pm.

19 mL of 0.50 mool/L NaOH which is standardized becomes titrated alongside 24 mL of 0.44 mol/L acetic acid. Determine the pH of the solution

Please judge my work:

Becasue NaOH and acetic acid react in a 1:1 ratio,

initital moles of acetic acid
=0.0019 L x 0.44M
= 0.000836 mols

mols of acetic acid neutralized is
= 0.0024 L x 0.44 M
=0.0019 L

molfs of acetic acid is 0.0019-0.000836 =0.001064

Whats my next steps?

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