Answer:
- Take 3.3 mL of 3.0-M hydrochloric acid and subsequently add 76.7 mL of water to complete the 100.00 mL.
- Take 11.7mL of 6.0-M hydrochloric acid and subsequently add 88.3 mL of water to complete the 100.00 mL
Explanation:
Hello,
In this case, given that the dilutions are preparedfrom 3.0-M and 6.0-M hydrochloric acid, we must proceed as follows:
- 3.0-M stock: when using this stock, the aliquot you must take is computed as shown below:
[tex]V_1=\frac{M_2V_2}{M_1}=\frac{100.00mL*0.70M}{3.0M}=23.3mL[/tex]
It means that you must take 23.3 mL of 3.0-M hydrochloric acid and subsequently add 76.7 mL of water to complete the 100.00 mL.
- 6.0-M stock: when using this stock, the aliquot you must take is computed as shown below:
[tex]V_1=\frac{M_2V_2}{M_1}=\frac{100.00mL*0.70M}{6.0M}=11.7mL[/tex]
It means that you must take 11.7mL of 6.0-M hydrochloric acid and subsequently add 88.3 mL of water to complete the 100.00 mL.
Regards.