calculate the ph of a solution by addition of 0.1 mL of 0.5m hcl to 50.0 ml pure water.

I'm supposed to use c1v1=c2v2...
but I did 0.1/50ml=0.002....
0.002x0.5.. then -log the answer and got 3 for a ph is it right?

yes, correct.

(0.002)0.5=

To calculate the pH of a solution using the given information, you need to consider the dissociation of the HCl (hydrochloric acid) in water. The initial volume and concentration of HCl are 0.1 mL and 0.5 M, respectively. The final volume of the solution after adding the HCl is 50.0 mL.

To apply the formula c1v1 = c2v2, we can rearrange it as (c2v2) / (v1 + v2) = c1, where:
- c1 is the initial concentration of the acid (0.5 M)
- v1 is the initial volume of the acid (0.1 mL)
- c2 is the final concentration after adding the acid to the water (unknown)
- v2 is the volume of pure water added (50.0 mL)

Using the formula, we have:
(c2 * 50.0 mL) / (0.1 mL + 50.0 mL) = 0.5 M

Simplifying the equation gives:
c2 / 50.1 mL = 0.5 M

Rearranging the equation to solve for c2, we find:
c2 = (0.5 M) * (50.1 mL)

Calculating c2, we get:
c2 = 25.05 M/mL

However, we cannot have a concentration expressed in M/mL. Instead, we need to convert the units of mL to L (liters) by dividing by 1000.

c2 = 25.05 M / (1000 mL/L)

c2 = 0.02505 M

Now that we have the final concentration, we can calculate the pH using the formula -log10[H+].

pH = -log10(0.02505)

Calculating the pH gives an approximate value of 1.6.

Therefore, the pH of the solution after adding 0.1 mL of 0.5 M HCl to 50.0 mL of pure water is approximately 1.6.