1) Suppose you have 1 liter of a solution thats 60% hydrogen chloride (HCl), i.e

Question

1) Suppose you have 1 liter of a solution thats 60% hydrogen chloride (HCl), i.e. a 50 % solution of hydrochloric acid. Calculate the amount of pure water you have to add to get the following concentrations. HCl concentrations 50%, 40%, 30%, and 20%.

https://gyazo.com/bd71dcf70b7556a7631f4975e88a5080

2) Suppose you want to get a solution whose concentration is c. How much water w will you need to add to get that concentration? To get this result, repeat your calculations from the previous question but use c for the final concentration. Your final result should be a formula with both a w and a c in it that you've solved for c, e.g. w=c+.06

Explanation / Answer

We have 60% HCl solution

1) We are preparing 50% HCl solution

(C1*V1)=(C2*V2)

(60*X) = (50*1000)

=833.33ml

= 1000-833.33

= 166.66 ML of pure water we need to prepare for 50% solution

2) 40% HCl solution

(C1*V1)=(C2*V2)

(60*X) = (40*1000)

=666.66ml

= 1000-666.66

= 333.33 ML of pure water we need to prepare for 40% solution

3) 30% HCl solution

(C1*V1)=(C2*V2)

(60*X) = (30*1000)

=500 ml

= 1000-500

= 500 ml of pure water we need to prepare for 30% solution

4) 20% HCl solution

(C1*V1)=(C2*V2)

(60*X) = (20*1000)

=333.33 ml

= 1000-333.33

= 666.66 ml of pure water we need to prepare for 20% solution

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.