Thank you. I'm diluting rooting hormone to soak green giant arborvitae tree cuttings overnight. Now I know how to dilute my stock solution. I will successfully root my tree cuttings and sell them next year. You helped me make money grow on trees. Money really does grow on trees if you propagate and sell the right ones. Propagate and sell screening plants, hedge plants and ground covers. People usually by a lot more than 1 of these type of plants,
Hello and thank you for your great videos , i have a question please : when diluting a weak acid i noticed in the book that they use the dilution formula :C1V1=C2V2 based on the fact that the number of moles will stay constant , however for weak acid we know that upon adding water the dissociation of the weak acid increases to maintain balance and hence the number of moles of the weak acid decreases , so based on that how the dilution formula still applied?...thank you in advance.
The number of moles of weak acid does not change upon dilution because the dissociation of the weak acid in water is a reversible process. As the weak acid dissociates in water, it produces hydronium ions and the conjugate base of the acid. The concentration of hydronium ions increases, which increases the acidity of the solution and causes the weak acid to dissociate less. This is known as the common ion effect. When the weak acid is diluted with water, the total number of moles of the acid remains constant, but the volume of the solution increases. This results in a decrease in the concentration of the weak acid, which in turn decreases the concentration of hydronium ions in the solution. As a result, the dissociation of the weak acid increases to maintain the equilibrium concentration of hydronium ions. However, the increase in dissociation upon dilution is small compared to the total amount of weak acid present, and the total number of moles of the weak acid remains essentially constant. The dilution equation, C1V1=C2V2, assumes that the number of moles of solute remains constant upon dilution, which is generally true for weak acids and bases.
Hello Adam, Greeting to you from the UK, Great video I tried to copy the formula but I'm still too slow to grab your simplified teaching may be for various reasons. I wish you can help me here. Ok here is confusion, I'm trying to make a fertilizer stock solution of Mn in ppm I need Mn 0.8 ppm, in my final solution I do have a conversion factor i.e 4.061. so 0.8 times 4.061 = 3.2488, I take out the element purity of 90% which will be 3.2 *100/90 = 3.6. So this means I must have 3.6mg/l if I'm not mistaken. Now I want to make a stock solution 600x concentrate in a 50 liter tank, so 3.6*600 =2160 to convert into grams I /(divide) by 1000, so 3.6*600/1000 = 2.16gram . Am I right until now ? If I'm right I go a little further where the confusion begins and I need your precious help So my final solution tank is 3785liter and my stock solution of 600x concentration is 37.85 liter so do I need to take the 2.16grams *(times) 37.85 liter stock tank volume to get the total salt dissolved to make a 600x concentration which is 2.16*37.85 =81.756 or 81.8grams of salt for a 600x concentrate if this is correct please guide me further on how much of stock solution to add to make a final solution for a 3785liter tank of 0.8ppm? Adam I highly appreciate your time and your efforts in getting this solved for me as if I understand this I can work on a few more salts, I'm feeling very embarrassed with my breakdown but I rather put it simple for your understanding and more over for my understanding. Alternatively you can send me a reply on sidneyspleno13@gmail.com I thank you very much for your time and do look forward of your e-teaching. Sidney Rodrigues.
IMPORTANT QUESTION: in what situation if ever are you able to solve for one of the two variables on the right side of the equation, that is, either c2 or v2?
you’re indeed a man of your words #Dr.Ani John on RUclips, after purchasing and using his herbal product for HSV2 I was totally cured permanently within few days. Thanks to #Dr.Ani John for the fast recovery
Hey. Unfortunately the dilution equation can be applied only on molar concentrations. See Skoog : Fundamentals of analytical chemistry, 9th edition. page 75. " This equation is based on the fact that the number of moles of solute in the diluted solution must be equal the number of moles in the concentrated reagent.' For example: Let's say that you want to prepare 1 L of 15% H2SO4 solution (density: 1.095 g/mL) from 60 % H2SO4 (density 1.498 g/mL). We have to find out the volume needed. If we use the dilution equation we get that we need 250 mL of 60% H2SO4, but if we calculate correctly we obtain that we actually need only 182.7 mL. Here is the correct calculation: 15% means that 15 g H2SO4 is in 100 g solution. Using its density we calculate the solution's volume, that is V=m/d=100/1.095=91.32 mL. Next we calculate the amount of pure H2SO4 that is in 1L=1000 mL solution, which is going to equal 164.25 g pure H2SO4. Than we say that 60 g H2SO4 is in .......100 g solution 164.25 g H2SO4.........x = 273.76 g solution. And finally we transform the obtained mass into volume using it's density and the formula given earlier, so we get 182.7 mL, instead of 250 mL that we would get using the dilution equation.
Thank you. I'm diluting rooting hormone to soak green giant arborvitae tree cuttings overnight. Now I know how to dilute my stock solution. I will successfully root my tree cuttings and sell them next year. You helped me make money grow on trees. Money really does grow on trees if you propagate and sell the right ones. Propagate and sell screening plants, hedge plants and ground covers. People usually by a lot more than 1 of these type of plants,
amazing! cleared out my doubt! thank u!
thank you this video helped me a a lot !
Bro thank you I had to make a dilution lab and I forgot how this works you saved my butt thank you!
In my school we take biology freshman year and chemistry sophomore lol
Thank you so much. Excellent simplified explanation.
Is this principle used in the manufacture of IV pharmaceutical solutions? They probably are since there's not much room for error
I just have trouble knowing what’s the c1 v1 and c2 v2 word problems confuse me 😥
Same problem
how to find c2/vs
clearly explained
Hello and thank you for your great videos , i have a question please : when diluting a weak acid i noticed in the book that they use the dilution formula :C1V1=C2V2 based on the fact that the number of moles will stay constant , however for weak acid we know that upon adding water the dissociation of the weak acid increases to maintain balance and hence the number of moles of the weak acid decreases , so based on that how the dilution formula still applied?...thank you in advance.
The number of moles of weak acid does not change upon dilution because the dissociation of the weak acid in water is a reversible process. As the weak acid dissociates in water, it produces hydronium ions and the conjugate base of the acid. The concentration of hydronium ions increases, which increases the acidity of the solution and causes the weak acid to dissociate less. This is known as the common ion effect.
When the weak acid is diluted with water, the total number of moles of the acid remains constant, but the volume of the solution increases. This results in a decrease in the concentration of the weak acid, which in turn decreases the concentration of hydronium ions in the solution. As a result, the dissociation of the weak acid increases to maintain the equilibrium concentration of hydronium ions.
However, the increase in dissociation upon dilution is small compared to the total amount of weak acid present, and the total number of moles of the weak acid remains essentially constant. The dilution equation, C1V1=C2V2, assumes that the number of moles of solute remains constant upon dilution, which is generally true for weak acids and bases.
Thanks
You said early that you can't cancel unit's but later you do
Hello Adam,
Greeting to you from the UK,
Great video I tried to copy the formula but I'm still too slow to grab your simplified teaching may be for various reasons.
I wish you can help me here.
Ok here is confusion, I'm trying to make a fertilizer stock solution of Mn in ppm
I need Mn 0.8 ppm, in my final solution I do have a conversion factor i.e 4.061. so 0.8 times 4.061 = 3.2488, I take out the element purity of 90% which will be 3.2 *100/90 = 3.6. So this means I must have 3.6mg/l if I'm not mistaken. Now I want to make a stock solution 600x concentrate in a 50 liter tank, so 3.6*600 =2160 to convert into grams I /(divide) by 1000, so 3.6*600/1000 = 2.16gram . Am I right until now ?
If I'm right I go a little further where the confusion begins and I need your precious help
So my final solution tank is 3785liter and my stock solution of 600x concentration is 37.85 liter so do I need to take the 2.16grams *(times) 37.85 liter stock tank volume to get the total salt dissolved to make a 600x concentration which is 2.16*37.85 =81.756 or 81.8grams of salt for a 600x concentrate if this is correct please guide me further on how much of stock solution to add to make a final solution for a 3785liter tank of 0.8ppm?
Adam I highly appreciate your time and your efforts in getting this solved for me as if I understand this I can work on a few more salts,
I'm feeling very embarrassed with my breakdown but I rather put it simple for your understanding and more over for my understanding.
Alternatively you can send me a reply on sidneyspleno13@gmail.com
I thank you very much for your time and do look forward of your e-teaching.
Sidney Rodrigues.
Sidney Rodrigues ik you tight because they didn’t respond
In our institute we take AP chemistry and I am just in grade 9-ASP( Advanced Science program)
IMPORTANT QUESTION: in what situation if ever are you able to solve for one of the two variables on the right side of the equation, that is, either c2 or v2?
Thank you
very good explanation
Thank you 🙏🙏🙏
you’re indeed a man of your words #Dr.Ani John on RUclips, after purchasing and using his herbal product for HSV2 I was totally cured permanently within few days. Thanks to #Dr.Ani John for the fast recovery
Hey. Unfortunately the dilution equation can be applied only on molar concentrations. See Skoog : Fundamentals of analytical chemistry, 9th edition. page 75. " This equation is based on the fact that the number of moles of solute in the diluted solution must be equal the number of moles in the concentrated reagent.'
For example: Let's say that you want to prepare 1 L of 15% H2SO4 solution (density: 1.095 g/mL) from 60 % H2SO4 (density 1.498 g/mL). We have to find out the volume needed.
If we use the dilution equation we get that we need 250 mL of 60% H2SO4, but if we calculate correctly we obtain that we actually need only 182.7 mL.
Here is the correct calculation:
15% means that 15 g H2SO4 is in 100 g solution. Using its density we calculate the solution's volume, that is V=m/d=100/1.095=91.32 mL.
Next we calculate the amount of pure H2SO4 that is in 1L=1000 mL solution, which is going to equal 164.25 g pure H2SO4.
Than we say that 60 g H2SO4 is in .......100 g solution
164.25 g H2SO4.........x = 273.76 g solution.
And finally we transform the obtained mass into volume using it's density and the formula given earlier, so we get 182.7 mL, instead of 250 mL that we would get using the dilution equation.
very true!