Question

You used 25 mL of 6.0 M NaOH to neutralize 100 mL of an unknown acid in a titration experiment. What is the molarity of the acid?

Answer 1

Answer: The molarity of the acid is 1.5 M.

Step-by-step explanation:

We can use the equation:

`M_1 V_1 = M_2V_2`

where M represents the molarity of the solution, and V represents the volume.

Then, plugging in our information, we have:

`(X)(100 \text{mL}) = (6.0 \text{ M})(25 \text{ mL})`

Solving for X, by dividing by 100, gives us:

`X = \frac{(6.0 \text{ M})(25 \text{ mL})}{100 \text{ mL}} = 1.5 \text{ M}`

So, the molarity of the unknown acid is 1.5 M.

Answer 2

c(H+)=1.5M that is your answer