Suppose you are titrating an unknown acid soluition with 0.100 N KOH. What is the normality of the acid if 25.00 mL of it requires 35.20 mL of base?

I got 0.07 N
I am not sure if I did the problem correctly and need some help.

Normality*25ml=.1*35.20

No, your answer is wrong, by a factor of 2...why is that?

I placed the 0.1 in the wrong part of the equation....

so the normality is 0.14

To find the normality of the unknown acid, you need to use the titration data provided. Here's how you can calculate it:

1. Start by writing the balanced chemical equation for the reaction between the acid and the base. For simplicity, let's assume the reaction is a 1:1 stoichiometry:

Acid (HA) + Base (BOH) → Salt (BA) + Water (H2O)

2. Determine the mole ratio between the acid and the base from the balanced equation. In this case, it is 1:1, meaning one mole of acid reacts with one mole of base.

3. Calculate the number of moles of KOH (base) used in the reaction. This can be done using the equation:

Moles of KOH = volume of KOH (L) × normality of KOH (N)

Given that the volume of KOH used is 35.20 mL = 0.03520 L and the normality of KOH is 0.100 N, you can substitute these values into the equation:

Moles of KOH = 0.03520 L × 0.100 N = 0.00352 moles of KOH

4. Since the mole ratio between the acid and base is 1:1, the number of moles of acid consumed in the reaction is also 0.00352 moles.

5. Calculate the concentration or normality of the acid using the equation:

Normality of acid = Moles of acid / volume of acid (L)

Given that the volume of acid used is 25.00 mL = 0.02500 L, you can substitute the values:

Normality of acid = 0.00352 moles / 0.02500 L = 0.1408 N

Therefore, the normality of the unknown acid is approximately 0.1408 N, not 0.07 N as you had initially calculated.