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, OK to here but check your numbers. They don't make sense.
initital moles of acetic acid
=0.0019 L x 0.44M It's 19 mL of NaOH which is 0.44 M. You've mixed up mL and component, I believe.
= 0.000836 mols
mols of acetic acid neutralized is
= 0.0024 L x 0.44 M The one above also says acaeitic acid so which is which?
=0.0019 L Liters???. You started out by saying this was mols. Is molarity measured in L?
molfs of acetic acid is 0.0019-0.000836 =0.001064
Whats my next steps?
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?
1 answer
1 answer