Question

How many milliliters (ml) of a 20% hydrochloric acid (HCI) solution must a chemist add to 813.75 ml of a 55% HCl solution to get a 51% HCl solution mixture? milliliters of 20% hydrochloric acid solution (NO COMMAS) Work area number of milliliters hydrochloric acid strength | Amount of hydrochloric acid 20% HCl solution 55% HCl solution 51% HCl solution Question Help: Message instructor

Answer 1

Given, percentage of HCl in first solution, R1=20%.

Percentage of HCl in second solution, R2=55%.

The amount of second solution, A2=813.75 ml.

The percentage of HCl in mixture, R=20%.

Let A1 be the amount of first solution added to get mixture.

Hence,

`(R)/(100)(A1+A2)=(R1)/(100)* A1+(R2)/(100)* A2`

Now put the values in above equation.

`\begin{gathered} (51)/(100)*(A1+813.75)=(20)/(100)* A1+(55)/(100)*813.75 \\ 51*(A1+813.75)=20* A1+55*813.75 \\ 51* A1+51*813.75=20* A1+55*813.75 \\ 51* A1-20* A1=55*813.75-51*813.75 \\ A1(51-20)=813.75(55-51) \\ A1*31=813.75*4 \\ A1=(813.75*4)/(31) \\ A1=105 \end{gathered}`

Therefore, 105 ml of 20% HCl is to be added to 813.75 ml of a 55% HCl to get 51% HCl solution mixture.