Question

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?

Answer 1

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, <b>OK to here but check your numbers. They don't make sense.</b>

initital moles of acetic acid
=0.0019 L x 0.44M <B>It's 19 mL of NaOH which is 0.44 M. You've mixed up mL and component, I believe.</b>
= 0.000836 mols

mols of acetic acid neutralized is
= 0.0024 L x 0.44 M <b>The one above also says acaeitic acid so which is which?</b>
=0.0019 L <b>Liters???. You started out by saying this was mols. Is molarity measured in L?</b>

molfs of acetic acid is 0.0019-0.000836 =0.001064

Whats my next steps?