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How would you make a 200mL solution of a 2:3 HCl solution?
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+ Chemistry
+ Hydrochloric acid
Posted by
Marco Pereira
How would you make a 200mL solution of a 2:3 HCl solution?
Quite simple :
Take a 150 cm³ volumetric flask ( Question : Is there such a piece of glassware? I doubt it - so take a measuring cylinder with capacity = 250 cm³ . )
Fill this with distilled water to approx 100 cm³ .
Now use a 10cm³ graduated pipette to measure out 9 cm³ of the concentrated acid ( CAREFUL!! - use proper suction equipment - not your mouth )
Add the 9 cm³ HCl to the 100 cm³ in the measuring cylinder .
Add further distilled water to final volume = 150 cm³ . Mix thoroughly with a glass stirring rod .
Note - In chemistry we do not produce analytical solutions of a certain total volume , by mixing two ( or more) intermediate volumes . In this case do not answer: Mix 9 cm³ of HCl with 141 cm³ distilled water
Take a 150 cm³ volumetric flask ( Question : Is there such a piece of glassware? I doubt it - so take a measuring cylinder with capacity = 250 cm³ . )
Fill this with distilled water to approx 100 cm³ .
Now use a 10cm³ graduated pipette to measure out 9 cm³ of the concentrated acid ( CAREFUL!! - use proper suction equipment - not your mouth )
Add the 9 cm³ HCl to the 100 cm³ in the measuring cylinder .
Add further distilled water to final volume = 150 cm³ . Mix thoroughly with a glass stirring rod .
Note - In chemistry we do not produce analytical solutions of a certain total volume , by mixing two ( or more) intermediate volumes . In this case do not answer: Mix 9 cm³ of HCl with 141 cm³ distilled water
Quite simple :
Take a 150 cm³ volumetric flask ( Question : Is there such a piece of glassware? I doubt it - so take a measuring cylinder with capacity = 250 cm³ . )
Fill this with distilled water to approx 100 cm³ .
Now use a 10cm³ graduated pipette to measure out 9 cm³ of the concentrated acid ( CAREFUL!! - use proper suction equipment - not your mouth )
Add the 9 cm³ HCl to the 100 cm³ in the measuring cylinder .
Add further distilled water to final volume = 150 cm³ . Mix thoroughly with a glass stirring rod .
Note - In chemistry we do not produce analytical solutions of a certain total volume , by mixing two ( or more) intermediate volumes . In this case do not answer: Mix 9 cm³ of HCl with 141 cm³ distilled water
Quite simple :
Take a 150 cm³ volumetric flask ( Question : Is there such a piece of glassware? I doubt it - so take a measuring cylinder with capacity = 250 cm³ . )
Fill this with distilled water to approx 100 cm³ .
Now use a 10cm³ graduated pipette to measure out 9 cm³ of the concentrated acid ( CAREFUL!! - use proper suction equipment - not your mouth )
Add the 9 cm³ HCl to the 100 cm³ in the measuring cylinder .
Add further distilled water to final volume = 150 cm³ . Mix thoroughly with a glass stirring rod .
Note - In chemistry we do not produce analytical solutions of a certain total volume , by mixing two ( or more) intermediate volumes . In this case do not answer: Mix 9 cm³ of HCl with 141 cm³ distilled water
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concentration of the acid???
this is a ratio mixing problemproblem
2/3 solution
2/5 = 40%
3/5 = 60%
///////////////////////////////////
200 x 0.4 = 80 mL HCl
300 x 0.6 = 120 mL H2O
mix these volumes.
(acid to water)
we do not know the concentration of the acid
////////////////////////////////////
concentration of the acid???
this is a ratio mixing problemproblem
2/3 solution
2/5 = 40%
3/5 = 60%
///////////////////////////////////
200 x 0.4 = 80 mL HCl
300 x 0.6 = 120 mL H2O
mix these volumes.
(acid to water)
we do not know the concentration of the acid
////////////////////////////////////
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