What volume of water must be added to 11.1 ml of a ph 2.0 solution of hno3 in order to change the ph to 4.0? a) 11.1 ml b) 89 ml c) 110 ml d) 1.10 ? 103 ml e) 28 ml

Volume of water that must be added is 1.10 L

Explanation:

pH measures the acidity or the alkalinity of a substance

It is given by;

pH = -log[H+]

Using this we can find the concentration of H+ ions in the acid

pH = 2 = -log[H+]

Therefore;

[H+] = 10^-2

       = 0.01 M

But, since 1 mole HNO₃ ionizes to give 1 mole of H+, then the concentration of HNO₃ is equal to the concentration of H+ ([HNO₃] = [H+])

Therefore;

Initial [HNO₃] = 0.01 M

Initial volume of HNO₃ = 11.1 mL or 0.0111 L

We can then use dilution equation to find the final volume after dilution.

The final pH is 4

Therefore, [H+] = 10^-4

                         = 0.0001 M

Thus, the final concentration of HNO₃ is 0.0001 M

Using dilution equation;

M1V1 =M2V2

Thus; V2 = M1V1÷ M2

               = (0.01 M× 0.0111 L)÷ 0.0001 M

               = 1.11 L

This means the final total volume will 1.11 L or 1110 ml

Therefore; The volume of water added = 1110 ml - 11.1 ml

                                                                  = 1098.9 ml or

                                                                  = 1.0989 L

                                                                  = 1.10 L(2 d.p.)

Hence, The volume of water that must be added is 1.10 L

im pretty sure its b