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,

$\frac{R}{100}(A1+A2)=\frac{R1}{100}\times A1+\frac{R2}{100}\times A2$

Now put the values in above equation.

$\begin{gathered} \frac{51}{100}\times(A1+813.75)=\frac{20}{100}\times A1+\frac{55}{100}\times813.75 \\ 51\times(A1+813.75)=20\times A1+55\times813.75 \\ 51\times A1+51\times813.75=20\times A1+55\times813.75 \\ 51\times A1-20\times A1=55\times813.75-51\times813.75 \\ A1(51-20)=813.75(55-51) \\ A1\times31=813.75\times4 \\ A1=\frac{813.75\times4}{31} \\ A1=105 \end{gathered}$

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