Simple Keq Acid/Base Reaction Question (1 Viewer)

[Pedro123]

Very sorry if this is a pretty stupid question, probably just a pre-exam freakout:

In a beaker, 50 mL of 1.0 mol L−1 CH3COOH is reacted with 30 mL of 0.70 mol L−1 NaOH. (Ka of CH3COOH = 1.7 x 10−5)
Calculate the pH of the final solution

My confusion is how are you supposed to calculate the pH of the solution after the neutralisation occurs. The acetic acid is in excess, meaning after the reaction there will be left over H30+ ions, but also ethanoate ions, in which case it doesn't make sense how you determine the pH of the solution, since you can't use the same "assume x is [H30+], as x is very small ...." (Since part of the equilibrium has already shifted, and the neutralisation will mean the equilibrium shifts to the right more)

Thanks

EDIT: For reference, the answer I keep getting is 4.6
n(NaOH) = 0.021
n(CH3COOH) = 0.05
n(CH3COOH) remaining = 0.05-0.021 = 0.029
At this point, n(CH3COOH) = 0.029, n(CH3COO-) = 0.021 (Frm neutralisation)
Converting to concentrations and putting them into Keq expression, (Making the assumption that from this point, ionisation degree is very low, therefore if [H3O] = x, then [CH3COOH] - x = [CH3COOH], and [CH3COO-] + x = [CH3COO-]
[H3O+] = 2.4x10^-5 (Justifying assumption)
=> pH = 4.6

 

[CM_Tutor]

Very sorry if this is a pretty stupid question, probably just a pre-exam freakout:

In a beaker, 50 mL of 1.0 mol L−1 CH3COOH is reacted with 30 mL of 0.70 mol L−1 NaOH. (Ka of CH3COOH = 1.7 x 10−5)
Calculate the pH of the final solution

My confusion is how are you supposed to calculate the pH of the solution after the neutralisation occurs. The acetic acid is in excess, meaning after the reaction there will be left over H30+ ions, but also ethanoate ions, in which case it doesn't make sense how you determine the pH of the solution, since you can't use the same "assume x is [H30+], as x is very small ...." (Since part of the equilibrium has already shifted, and the neutralisation will mean the equilibrium shifts to the right more)

Thanks

EDIT: For reference, the answer I keep getting is 4.6
n(NaOH) = 0.021
n(CH3COOH) = 0.05
n(CH3COOH) remaining = 0.05-0.021 = 0.029
At this point, n(CH3COOH) = 0.029, n(CH3COO-) = 0.021 (Frm neutralisation)
Converting to concentrations and putting them into Keq expression, (Making the assumption that from this point, ionisation degree is very low, therefore if [H3O] = x, then [CH3COOH] - x = [CH3COOH], and [CH3COO-] + x = [CH3COO-]
[H3O+] = 2.4x10^-5 (Justifying assumption)
=> pH = 4.6

Find the concentrations of acetic acid and acetate in the solution from the neutralisation, then put those values as knowns into your K_a - the concentration of the hydronium will be the only unknown.