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?
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?
Your work so far is correct. The next step is to calculate the concentration of acetic acid after neutralization, and from that, determine the pH of the solution.
To find the concentration of acetic acid, divide the number of moles of acetic acid by the total volume of the solution:
Concentration of acetic acid = moles of acetic acid / total volume of solution
moles of acetic acid = 0.001064 mol (as you calculated)
total volume of solution = the sum of the volumes of NaOH and acetic acid = 19 mL + 24 mL = 43 mL = 0.043 L
Concentration of acetic acid = 0.001064 mol / 0.043 L
≈ 0.0248 mol/L
Now, to determine the pH of the solution, we need to find the concentration of hydronium ions (H3O+), which comes from the dissociation of acetic acid. Acetic acid is a weak acid and only partially dissociates in water. The dissociation equation is:
CH3COOH + H2O ⇌ CH3COO- + H3O+
The equilibrium expression for this dissociation is given by:
Ka = [CH3COO-][H3O+] / [CH3COOH]
The Ka value for acetic acid is known to be approximately 1.8 x 10^-5.
Since acetic acid and the acetate ion have a 1:1 stoichiometry, we can assume that [CH3COO-] is equal to the concentration of acetic acid.
[H3O+] = (Ka * [CH3COOH]) / [CH3COO-]
= (1.8 x 10^-5 * 0.0248 mol/L) / 0.0248 mol/L
≈ 1.8 x 10^-5
Finally, to find the pH, we can take the negative logarithm (base 10) of the concentration of hydronium ions:
pH = -log[H3O+]
= -log(1.8 x 10^-5)
≈ 4.74
Therefore, the pH of the solution is approximately 4.74.