A solution of HCL whose pH=2.010 was added to neautralize a 40.00 ml solution of KOH, the molarity of KOH is .2567 M. How many mls of HCl were required?

To solve this problem, we can use the concept of neutralization and the equation of molarity:

Molarity (M) = moles of solute / liters of solution

Let's calculate the moles of KOH in the 40.00 ml solution:

Moles of KOH = Molarity of KOH × Volume of KOH solution
= 0.2567 M × 0.04000 L
= 0.010268 mol

Since the HCl solution is used to neutralize the KOH solution, the moles of HCl required will be equal to the moles of KOH:

Moles of HCl = Moles of KOH
= 0.010268 mol

Now, we need to calculate the volume of HCl solution.

Rearrange the molarity equation to calculate the volume:

Volume of solution (L) = Moles of solute / Molarity

Volume of HCl solution = 0.010268 mol / Molarity of HCl

Given that the pH of the HCl solution is 2.010, we can convert the pH to the concentration of H+ ions to determine the molarity of HCl.

pH = -log10[H+]

Rearrange the equation to calculate the concentration of H+ ions:

[H+] = 10^(-pH)

[HCl] = [H+]

[HCl] = 10^(-2.010) M

Now, substitute the value of molarity of HCl:

Volume of HCl = 0.010268 mol / (10^(-2.010) M)

Calculate the volume:

Volume of HCl = 0.010268 mol / (10^(-2.010))

Volume of HCl = 0.010268 mol / 0.008340096

Volume of HCl = 1.231 ml

Therefore, approximately 1.231 ml of HCl solution are required to neutralize the 40.00 ml solution of KOH.