how many mL of .100 M of acetic acid and .100 M of NaOH, would be need to make the volume 1.0 liters and have a pH7.00 if Ka=1.8x10^-5?

can someone please help me with steps to get the answer

Use the Henderson-Hasselbalch equation.

can you elaborate a little more on how to so it

To find out how many mL of acetic acid and NaOH are needed to make a solution of 1.0 liter with a pH of 7.00, and given that the Ka value for acetic acid is 1.8x10^-5, follow the steps below:

Step 1: Determine the balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):
CH3COOH + NaOH → CH3COONa + H2O

Step 2: Write the equation for the ionization of acetic acid in water:
CH3COOH ⇌ CH3COO- + H+

Step 3: Calculate the initial concentration of acetic acid (CH3COOH) and sodium hydroxide (NaOH) required to obtain a pH of 7.00:

- Since the volume is given as 1.0 liter, the initial concentration of both acetic acid and NaOH will be the same.
- The pH of a solution is related to the concentration of H+ ions by the formula: pH = -log[H+]
- Since pH = 7.00, [H+] = 10^(-7.00) = 1.0 × 10^(-7) M

Step 4: Use the Ka expression for acetic acid (CH3COOH) to calculate the initial concentration of the acetate ion (CH3COO-) in terms of [H+]:
Ka = [CH3COO-][H+] / [CH3COOH]

1.8 × 10^(-5) = [CH3COO-][1.0 × 10^(-7)] / [CH3COOH]

Step 5: Since the initial concentration of acetic acid (CH3COOH) is equal to the initial concentration of sodium hydroxide (NaOH) when they react completely, calculate the initial concentration of acetic acid using the equation obtained in step 4:
[CH3COOH] = [CH3COO-][H+] / Ka

[CH3COOH] = (1.0 × 10^(-7))(1.8 × 10^(-5)) / (1.8 × 10^(-5))

Step 6: Convert the initial concentration of acetic acid from moles per liter (M) to milliliters (mL):
[CH3COOH] = [CH3COOH] × volume (in mL) ÷ 1000

Volume (in mL) = [CH3COOH] × 1000 ÷ [CH3COOH]

Step 7: Substitute the value of [CH3COOH] from step 5 into the equation obtained in step 6 to calculate the required volume of acetic acid and NaOH in mL.

Remember to repeat steps 5 and 6 for NaOH as they have equal initial concentrations.

Please note that the above steps assume complete ionization of acetic acid and NaOH in water.