In a titration 25.1ml of a monoprotic weak acid is neutralized with 25.1 ml of .502M NaOH(aq).

Calculate the Ratio of [A-] to [HA] after 12.55 of the .502 M NaOH has been added to the initial amount of HA.

Anyone have any idea what's going on. I got as far as to find the mol's of NaOH which is .0063001 with 12.55mL at a concentration of .502

To calculate the ratio of [A-] to [HA] after 12.55 mL of 0.502 M NaOH has been added to the initial amount of HA, you need to determine the moles of both NaOH and the weak acid (HA). Here's how you can do that:

1. Convert the volume of NaOH added (12.55 mL) to liters: 12.55 mL * (1 L / 1000 mL) = 0.01255 L.

2. Now, calculate the moles of NaOH that was added:
Moles of NaOH = Volume (L) * Concentration (M)
Moles of NaOH = 0.01255 L * 0.502 M = 0.0063011 moles

3. Since the reaction is 1:1 between NaOH and HA, the number of moles of HA present is also 0.0063011 moles.

4. To find the ratio of [A-] to [HA], you need to determine the relative amounts in terms of moles. The weak acid (HA) can be represented as:
HA -> H+ + A-

The initial amount of HA before any reaction occurred is 0.0063011 moles. Since the moles of A- and HA are equal, the ratio between them is 1:1.

Therefore, the ratio of [A-] to [HA] after 12.55 mL of 0.502 M NaOH has been added to the initial amount of HA is 1:1.