Funghiarrosto

Hello everyone, I'm back again with another question, this time about buffer solutions. This assignment is divided into two parts, I've done the first one and I am quite confident in my answers. As for the second one, I can't see where the problem is in my calculations, but the results do not seem right to me. Here is the assignment: the objective is to make a buffer solution of 1L with pH = 8,50. Starting from 0,01mol KCN, how many moles of HCl need to be added in order to obtain the desired buffer solution, knowing that Ka of HCN is 4,8 x 10-10? What will be the variation of pH after adding 0,00005 moles of HClO4 to 100ml of buffer solution? Now as I said I think I have understood the first part. The problem with the second question is that the variation of pH I obtained is 3,12, but to me that seems wrong as, considering we are talking about buffer solution, the variation should be minimal. What have I done wrong? Here's my attempt: Molarity KCN = moles KCN / Volume Molarity KCN = 0,01mol/1L = 0,01mol/L (which from now on I'll call Cb, concentration of the base) I then calculated pKa for HCN: pKa HCN = -logKa pKa HCN = -log (4.8 x 10-10) pKa HCN = 9.32 I then used Henderson-Hasselbach equation: pH = pKa + log [base]/[acid] (or pH = pKa + log Cb/Ca) And I modified the equation to find log[base]/[acid]: log[base]/[acid] = pH1 -pKa log[base]/[acid] = 8,50 - 9,32 = -0,82 [base]/[acid] = 10-0,82 = 0,15 Ca = Cb/0,15 Ca = 0,01mol/L/0,15 = 0,06mol/L number of moles of HCl = 0,06mol/L x 1L = 0,06mol. This is where the first part ends and the second begins Molarity HClO4 ( = Ca) = 0,00005mol/0,1L = 0,0005mol/L Molarity KCN = 0,01mol/0,1L = 0,1mol/L I then used Henderson-Hasselbach again: pH = pKa + log[base]/[acid] pH2 = 9,32 + log(0,1mol/L / 0,0005mol/L) pH2 = 9,32 + 2,30 = 11,62 I finally calculated delta pH: pH2 - pH1 = 11,62 - 8,50 = 3,12, which I think is wrong because the variation is too great for a buffer solution. Any ideas as to where I went wrong??

Thank you very much for these valuable resources - I have a much clearer understanding of this subject now. I'll try to implement that formula in all of my future excercises. And thank you to both of you for correcting my mistakes! Without your help I'd still be stuck on that problem

It helped massively - thank you. That was such a dumb mystake on my part, I considered those 215mL of ethanol 70% as if they were milliliters of pure alcohol. It's all a lot clearer now. I rewrote the problem and corrected those mistakes. This should be correct, right? MY RESOLUTION: the purity of the alcohol I used is equal to 70%, meaning that in 100mL of product, there are 70mL of ethyl alcohol. With the proportion x mL : 100 = 100mL : 70, I obtained 134mL of ethyl alcohol with 70% purity, which is the equivalent of 100mL of ethyl alcohol with 100% purity. I then calculated the amount of ethyl alcohol that needs to be added to the solution: 30% of 500mL = 150mL of 100% pure ethyl alcohol. If 150mL of pure ethyl alcohol are necessary to obtain a solution with the desired concentration (30% v/v), then 1,5 x 143mL of ethyl alcohol pure at 70% will be necessary to obtain an aqueous solution of ethyl alcohol at 30% (v/v). I then calculated the amount of water necessary to reach 500mL of total solution: 500mL – 215mL (alcohol 70%) = 285mL of water. I then calculated the molecular weight of ethyl alcohol C2H5OH: C : 12,011g/mol H: 1,0079g/mol O: 15,9994g/mol Molecular weight: 2(12,001g/mol) + 6(1,0079g/mol) + 15,9994g/mol = 46,046g/mol. This is where the correction comes in: By adding 215mL of alcohol 70%, I get the concentration required of alcohol (30% v/v). That is because 70% of 215mL of non-pure ethyl alcohol are equal to 150mL of pure ethanol. So from now on I’ll use the values of pure alcohol. Density of pure ethyl alcohol: 789kg/m3 = 0,789g/cm3. Mass (g) = Volume (mL) x Density (g/cm3) Mass = 150mL x 0,789g/cm3 = 118,35g (total weight of PURE ethyl alcohol in the aqueous solution) 118,35g : 46,068g/mol = 2,5690 mol of ethyl alcohol. Molar concentration: 2,5690mol : 0,5L = 5,138M. For the molar concentration, you said I need to double the amounts so I can calculate it over 1L, so it’d be 5,138mol / 1L = 5,138M. The result is still the same, do I need to double the amounts because molarity must be calculated over 1L of solution, or is it still correct if I calculate it with the actual values I got in the problem?? To obtain 100mL of solution diluted 1:10, I took 1 part of concentrated solution and added 9 parts of diluent: (100mL : 10) x 1 = 10mL of initial concentrated solution. I then added water to these 10mL of solution to get to the desired volume of 100mL. For the second dilution, I took 10mL of the previous 1:10 dilution and added once again diluent to get to 100mL (so I just repeated the process, but on the 1:10 solution).

Thank you very much for your answer! I still do not understand why I need to refer to 150ml ethanol though.. the alcohol I'm using is not pure, so why shouldn't I use 215mL of ethanol 70%, which is the equivalent of 150mL at 100% purity? as for number 2), what I did was taking 1part of solution and added 9 parts diluent, not 10. So 1 part solute + 9 parts solvant = 10 parts in total which is the whole solution. In the end I'll have 1part solute : 10parts solution (1part solute + 9parts solvant). Is it wrong??

Thank you for your help! 500ml solution x .3 ethyl alcohol = 150ml alcohol (100% pure) is the correct proportion for pure ethanol, right? The one I'm using is at 70%, not 100% purity, that's why I looked up the density of 70% ethyl alcohol and used it to calculate its mass. How can I apply the operation 500ml x .3 ethyl alcohol = 150ml ethyl alcohol to my case??

Hi all, I'm a first year biology student at university, and I've only recently started my chemistry classes. I was given an assignment on solutions and dilutions, but I'm unsure about the results I've obtained... I'm not sure if the passages I've done are correct or if this could have been done in a simpler way. Could anyone point me towards the right direction and help me correct my mistakes? Here is the assignment with my resolution and my thought process: Prepare 500mL of an aqueous solution of ethyl alcohol at 30% (v/v) and determine its molar concentration. Then, from this solution, prepare: 100mL of solution diluted 1:10 100mL of solution diluted 1:100. MY RESOLUTION: the purity of the alcohol I used is equal to 70%, meaning that in 100mL of product, there are 70mL of ethyl alcohol. With the proportion x mL : 100 = 100mL : 70, I obtained 134mL of ethyl alcohol with 70% purity, which is the equivalent of 100mL of ethyl alcohol with 100% purity. I then calculated the amount of ethyl alcohol that needs to be added to the solution: 30% of 500mL = 150mL of 100% pure ethyl alcohol. If 150mL of pure ethyl alcohol are necessary to obtain a solution with the desired concentration (30% v/v), then 1,5 x 143mL of ethyl alcohol pure at 70% will be necessary to obtain an aqueous solution of ethyl alcohol at 30% (v/v). I then calculated the amount of water necessary to reach 500mL of total solution (I did this because I also need to make this experiment at home, so I know how much water I need to add. Now that I think about it, I probably could have skipped this part and just poured water until the solution reached 500mL): 500mL – 215mL (alcohol) = 285mL of water. I then calculated the molecular weight of ethyl alcohol C2H5OH: 😄 12,011g/mol H: 1,0079g/mol O: 15,9994g/mol Molecular weight: 2(12,001g/mol) + 6(1,0079g/mol) + 15,9994g/mol = 46,046g/mol. Density of ethyl alcohol with 70% purity: 885,56kg/m3 = 0,88556g/cm3. (found this online) Mass (g) = Volume (mL) x Density (g/cm3) Mass = 215mL x 0,88556g/cm3 = 190,40g (total weight of ethyl alcohol in the aqueous solution) 190,40g : 46,068g/mol = 4,1130 mol of ethyl alcohol. Molar concentration: 4,1150mol : 0,5L = 8,226M. (final answer to the first part of the problem) To obtain 100mL of solution diluted 1:10, I took 1 part of concentrated solution and added 9 parts of diluent: (100mL : 10) x 1 = 10mL of initial concentrated solution. I then added water to these 10mL of solution to get to the desired volume of 100mL. For the second dilution, I took 10mL of the previous 1:10 dilution and added once again diluent to get to 100mL (so I just repeated the process, but on the 1:10 solution). Is any of this correct? am I at least on the right path? I tried to compare results with some of my classmates, but we all got different answers 😕 Any form of help would really be greatly appreciated!!