Question

how much water should be added to 70 ml of 45 percent acid solution to dilute it to a 30 percent acid solution?

Answer 1

Chemistry => Solutions => Dilution

A solution is a mixture of two or more substances. In this case, we have a solution of acid in water. For dilutions we can apply the following equation:

`C_1V_1=C_2V_2`

Where,

C1 is the initial concentration, 45%

V1 is the initial volume of the solution, 70mL

C2 is the final concentration, 30%

V2 is the final volume of the solution, unknown

Now, we clear V2 and replace the known data:

`\begin{gathered} V_2=(C_1V_1)/(C_2) \\ \\ V_2=(45\%*70mL)/(30\%) \\ \\ V_2=105mL \end{gathered}`

The final volume of the solution will be 105 mL. The water should be added will be:

`\begin{gathered} H_2O=V_2-V_1 \\ \\ H_2O=105mL-70mL \\ \\ H_2O=35mL \end{gathered}`

Answer: To dilute the solution from 45% to 30%, should be added 35 mL of water.