Stoichiometry with Molar Concentration Molar Concentration Recall that
![Stoichiometry with Molar Concentration Stoichiometry with Molar Concentration](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-1.jpg)
![Molar Concentration • Recall that molar concentration is – c = n/v – We Molar Concentration • Recall that molar concentration is – c = n/v – We](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-2.jpg)
![Note! • If a volume is mentioned and the problem is about molarity, do Note! • If a volume is mentioned and the problem is about molarity, do](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-3.jpg)
![Example - 1 • Barium hydroxide has a mass of 0. 500 g. What Example - 1 • Barium hydroxide has a mass of 0. 500 g. What](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-4.jpg)
![Example - 2 • What volume of CO 2(g) at STP is produced if Example - 2 • What volume of CO 2(g) at STP is produced if](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-5.jpg)
![Titration • It is a process we use in chemistry to find the concentration Titration • It is a process we use in chemistry to find the concentration](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-6.jpg)
![](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-7.jpg)
![Titration • What we do – We have an unknown substance in an Erlenmeyer Titration • What we do – We have an unknown substance in an Erlenmeyer](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-8.jpg)
![Equivalence point or stoichiometric point • This is the point in a titration where Equivalence point or stoichiometric point • This is the point in a titration where](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-9.jpg)
![Example - 3 • H 2 SO 4 + 2 Na. OH → Na Example - 3 • H 2 SO 4 + 2 Na. OH → Na](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-10.jpg)
![Note! • Equivalence point is met when the moles are equal based on their Note! • Equivalence point is met when the moles are equal based on their](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-11.jpg)
![Example - 4 • H 2 SO 4 + 2 Na. OH → Na Example - 4 • H 2 SO 4 + 2 Na. OH → Na](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-12.jpg)
![Example - 5 • What volume of 0. 450 M KOH is required to Example - 5 • What volume of 0. 450 M KOH is required to](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-13.jpg)
![Remember! • In a titration, the equivalence point is when the molar ratios are Remember! • In a titration, the equivalence point is when the molar ratios are](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-14.jpg)
![Practice - 1 • Page 131 - #17 -25 Practice - 1 • Page 131 - #17 -25](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-15.jpg)
- Slides: 15
![Stoichiometry with Molar Concentration Stoichiometry with Molar Concentration](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-1.jpg)
Stoichiometry with Molar Concentration
![Molar Concentration Recall that molar concentration is c nv We Molar Concentration • Recall that molar concentration is – c = n/v – We](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-2.jpg)
Molar Concentration • Recall that molar concentration is – c = n/v – We can find moles by rearranging the equation to be – n = c*v – So we can find the number of moles if we know the concentration and the volume of the substance
![Note If a volume is mentioned and the problem is about molarity do Note! • If a volume is mentioned and the problem is about molarity, do](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-3.jpg)
Note! • If a volume is mentioned and the problem is about molarity, do not assume 22. 4 L as that is only for GAS at STP.
![Example 1 Barium hydroxide has a mass of 0 500 g What Example - 1 • Barium hydroxide has a mass of 0. 500 g. What](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-4.jpg)
Example - 1 • Barium hydroxide has a mass of 0. 500 g. What volume of HCl that is 0. 250 M is neutralized with this barium hydroxide? – Ba(OH)2 + 2 HCl → Ba. Cl 2 + 2 H 2 O • Neutralize for acid and base means the moles of acid = moles of base
![Example 2 What volume of CO 2g at STP is produced if Example - 2 • What volume of CO 2(g) at STP is produced if](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-5.jpg)
Example - 2 • What volume of CO 2(g) at STP is produced if 2. 5 L of 0. 150 M HCl reacts with excess Ca. CO 3?
![Titration It is a process we use in chemistry to find the concentration Titration • It is a process we use in chemistry to find the concentration](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-6.jpg)
Titration • It is a process we use in chemistry to find the concentration of an acid or base • We can perform this process when we know the exact volume and concentration of an acid or base and the volume of the substance we are neutralizing it • We use a tool called a burette to perform this as it is highly accurate and easy to control the volume down to 0. 1 m. L
![](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-7.jpg)
![Titration What we do We have an unknown substance in an Erlenmeyer Titration • What we do – We have an unknown substance in an Erlenmeyer](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-8.jpg)
Titration • What we do – We have an unknown substance in an Erlenmeyer flask with a known volume – We add our known concentration substance into the burette – We add an indicator into our unknown substance • Phenolphthalein – indicator turns pink at around p. H of 8. 5 – We slowly add or known substance into the flask and stop when our Erlenmeyer flask of solution is pink • This is our equivalence point
![Equivalence point or stoichiometric point This is the point in a titration where Equivalence point or stoichiometric point • This is the point in a titration where](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-9.jpg)
Equivalence point or stoichiometric point • This is the point in a titration where the moles of the acid and the moles of base meets the molar ratio of the balanced reaction – Their mole ratio matches the coefficients – Ba(OH)2 + 2 HCl → Ba. Cl 2 + 2 H 2 O – So the mole ratio will match when you have twice as many moles of HCl present compared to Ba(OH)2 as their ratio is 2: 1 and this is the equivalence point
![Example 3 H 2 SO 4 2 Na OH Na Example - 3 • H 2 SO 4 + 2 Na. OH → Na](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-10.jpg)
Example - 3 • H 2 SO 4 + 2 Na. OH → Na 2 SO 4 + 2 H 2 O – If an unknown concentration of H 2 SO 4 is in a flask and we add 0. 800 mol of Na. OH for it to reach the equivalence point, how many moles of H 2 SO 4 was in the flask? – If an unknown concentration of 0. 025 L Na. OH is in a flask and we add 0. 500 mol of H 2 SO 4. What is the concentration of Na. OH in the flask?
![Note Equivalence point is met when the moles are equal based on their Note! • Equivalence point is met when the moles are equal based on their](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-11.jpg)
Note! • Equivalence point is met when the moles are equal based on their molar ratio, not their concentration!
![Example 4 H 2 SO 4 2 Na OH Na Example - 4 • H 2 SO 4 + 2 Na. OH → Na](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-12.jpg)
Example - 4 • H 2 SO 4 + 2 Na. OH → Na 2 SO 4 + 2 H 2 O – It takes 21. 55 m. L of Na. OH at 0. 500 M to titrate 25. 00 m. L of H 2 SO 4. What is the molarity of H 2 SO 4 present?
![Example 5 What volume of 0 450 M KOH is required to Example - 5 • What volume of 0. 450 M KOH is required to](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-13.jpg)
Example - 5 • What volume of 0. 450 M KOH is required to react with 150 m. L of 0. 300 M H 3 PO 4 in order to produce a solution of K 2 HPO 4? – H 3 PO 4 + 2 KOH → K 2 HPO 4 + 2 H 2 O
![Remember In a titration the equivalence point is when the molar ratios are Remember! • In a titration, the equivalence point is when the molar ratios are](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-14.jpg)
Remember! • In a titration, the equivalence point is when the molar ratios are met thus it is the comparison between moles, not concentration! • 3 A + 2 B → 3 C – The equivalence point is met when we have any ratio of 3 A to 2 B.
![Practice 1 Page 131 17 25 Practice - 1 • Page 131 - #17 -25](https://slidetodoc.com/presentation_image_h/319f3c86696a6e903b49295522ac21d0/image-15.jpg)
Practice - 1 • Page 131 - #17 -25
Stoichiometry molar mass
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Molarity
Molarity image
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