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