What volume of 2.5% (m/v) koh can be prepared from 125 ml of a 5.0% koh solution?

Hey there!:

According to lew of dilution :

M*V = M'*V'

Where :

M = concentration of stock solution =  5.0% ( m / V )

V = volume of stock solution = 125 mL

M' = concentration of dilute solution = 2.5% ( m/V)

V' = volume of dilute solution = ??

Plug the values  we get:

V' =  m*V / M'

V' =  5.0 * 125 / 2.5

V' = 625 / 2.5

V' = 250 mL

Therefore , the volume prepared is 250 mL

Hope that helps!

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

i think that the answer is c