What volume of 11.80 mol/dm3 HCl needed to make 1 dm3 of 0.5 mol/dm3 aqueous HCl .

0.042dm3

Explanation:

Data obtained from the question include :

V1 (volume of stock solution) =?

C1 (concentration of stock solution) = 11.8mol/dm3

V2 (volume of the diluted solution) = 1 dm3

C2 (concentration of the diluted solution) = 0.5mol/dm3

Using the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:

C1V1 = C2V2

11.8 x V1 = 0.5 x 1

Divide both side by 11.8

V1 = 0.5/11.8

V1 = 0.042dm3

Therefore, 0.042dm3 of HCl is needed.

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.