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ammonia buffer solution
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Question:what is the volume of acid and base needed to mex (assume volume 1L)
Answers:In general, preparing a buffer solution requires either: A weak acid and a salt of the acid's conjugate base Or a weak base and a salt of the base's conjugate acid Both of which in sufficient amounts to maintain the ability to buffer now as to computation, i forgot my formula. LOL
Answers:In general, preparing a buffer solution requires either: A weak acid and a salt of the acid's conjugate base Or a weak base and a salt of the base's conjugate acid Both of which in sufficient amounts to maintain the ability to buffer now as to computation, i forgot my formula. LOL
Question:What volume of 0.24M HCl solution must be added to 227mL of 0.18 M ammonia solution to give a buffer solution with pH 8.99? (Kb ammonia = 1.8 * 10^5)
I got 0.417 L, not sure if that's correct. I just plugged everything into the buffer equation...but I think it only takes into account of the acid and its conjugate base
Help would be well appreciated. Thanks. Okay, I calculated this a different way and got 0.161 L
Answers:The assumption made here that the ammonia is neutralized. If this is the case, simply figure the amount of acid needed to neutralize the base. Start with a balanced chemical equation describing the reaction: HCl + NH4OH > NH4Cl + H2O Now, figure the moles of base present: (227 ml)*(1liter/1000 ml)*(0.18 moles/liter) = 0.041 moles The molar ratios from the chemical equation state the acid and base are on a 1:1 basis. Therefore, 0.041 moles of acid will be needed: (X ml)*(1 liter/1000 ml)*(0.24 moles/liter) = 0.041 moles X ml = 0.041 * 1000) / 0.24 = 9.80 ml Figuring the pH of this solution can be tricky. Ordinarily an ammonia solution has a Kb defined to be: Kb = [NH4(+)][OH()]/[NH4OH] In this case, all the ammonium ions are from the ammonium salt, which dissaossiates completely. There is only a small amount of either hydroxyl ions or ammonia molecules present. If "X" represents hydroxyl ions and ammonia molecules, then the equation becomes: 1.8E5 = [0.041 X][X]/[X] This will become an ordinary second degree quadratic equation of the form AX^2 + BX + C = 0 after a bit of algebra. The real root is the concentration of hydroxyl ions (the imaginary root will contain the square root of a negative number and can be discarded). From this, figure the pH as 14 + LOG([OH()]. If the acid completely neutralized the base, the pH ought to be 8.99.
Answers:The assumption made here that the ammonia is neutralized. If this is the case, simply figure the amount of acid needed to neutralize the base. Start with a balanced chemical equation describing the reaction: HCl + NH4OH > NH4Cl + H2O Now, figure the moles of base present: (227 ml)*(1liter/1000 ml)*(0.18 moles/liter) = 0.041 moles The molar ratios from the chemical equation state the acid and base are on a 1:1 basis. Therefore, 0.041 moles of acid will be needed: (X ml)*(1 liter/1000 ml)*(0.24 moles/liter) = 0.041 moles X ml = 0.041 * 1000) / 0.24 = 9.80 ml Figuring the pH of this solution can be tricky. Ordinarily an ammonia solution has a Kb defined to be: Kb = [NH4(+)][OH()]/[NH4OH] In this case, all the ammonium ions are from the ammonium salt, which dissaossiates completely. There is only a small amount of either hydroxyl ions or ammonia molecules present. If "X" represents hydroxyl ions and ammonia molecules, then the equation becomes: 1.8E5 = [0.041 X][X]/[X] This will become an ordinary second degree quadratic equation of the form AX^2 + BX + C = 0 after a bit of algebra. The real root is the concentration of hydroxyl ions (the imaginary root will contain the square root of a negative number and can be discarded). From this, figure the pH as 14 + LOG([OH()]. If the acid completely neutralized the base, the pH ought to be 8.99.
Question:How many grams of dry NH4Cl need to be added to 2.20 L of a 0.100 M solution of ammonia ,NH3, to prepare a buffer solution that has a pH of 8.85? Kb for ammonia is 1.8*10^5.
Answers:pKb = 4.7 pOH = 5.15 pKb = pOH + log [NH4+]/ [NH3] 4.7  5.15 =  0.45 10^0.45=0.355 = [NH4+]/ 0.100 [NH4+]= 0.0355 M moles = 0.0355 x 2.20 L=0.0781 Mass = 0.0781 x 53.49 g/mol= 4.18 g
Answers:pKb = 4.7 pOH = 5.15 pKb = pOH + log [NH4+]/ [NH3] 4.7  5.15 =  0.45 10^0.45=0.355 = [NH4+]/ 0.100 [NH4+]= 0.0355 M moles = 0.0355 x 2.20 L=0.0781 Mass = 0.0781 x 53.49 g/mol= 4.18 g
Question:How do I derive the buffer equation of an acetate/acetic acid buffer expressed explicitly in pH and an ammonia/ammonium buffer expressed explicitly in pOH?
Answers:You want to derive the Henderson  Hasselbalch equation  that is the name for the equation used to solve pH of buffer solutions. Because we are dealing with logarithms it is probably a good idea to refresh some of the basic log rules: log (A*B = log A + logB log (A/B) = logA  logB Log(A)^b = b*log A We always work with log base 10 in these problems. We will use these rules in the deriving of the HH equation: Consider a weak acid HA , which means that in water HA does not dissociate to a high degree. In practice we will assume that the acid dissociates less than 5%. This enables us to approximate that the molar concentration of the undissociated acid is the same as the original molar concentrtaion of the acid. HA will dissociate: HA H+ + A Ka = [H+] *[A] / [HA] In chemistry, if we add a p onto a term, it indicates: calculate the log of that term: Apply this to the Ka expression: log Ka = log ([H+]*[A+]/[HA]) log Ka = pKa pKa = log ([H+]*[A]/[HA]) Now refer to the log rules above, to rewrite the right hand side: pKa = log[A]  log[H+]  (log[HA]) You know that pH = log [H+], so substitute in the above: pKa = log[A] +pH +log[HA] Rearrange so that pH is the subject of the equation: pH = pKa + log [A]  log [HA] Log rules state that the expression (log[HA]) means divide by [HA] Rewrite as: pH = pKa + log ([A]/[HA]) That is how to derive the HH equation.
Answers:You want to derive the Henderson  Hasselbalch equation  that is the name for the equation used to solve pH of buffer solutions. Because we are dealing with logarithms it is probably a good idea to refresh some of the basic log rules: log (A*B = log A + logB log (A/B) = logA  logB Log(A)^b = b*log A We always work with log base 10 in these problems. We will use these rules in the deriving of the HH equation: Consider a weak acid HA , which means that in water HA does not dissociate to a high degree. In practice we will assume that the acid dissociates less than 5%. This enables us to approximate that the molar concentration of the undissociated acid is the same as the original molar concentrtaion of the acid. HA will dissociate: HA H+ + A Ka = [H+] *[A] / [HA] In chemistry, if we add a p onto a term, it indicates: calculate the log of that term: Apply this to the Ka expression: log Ka = log ([H+]*[A+]/[HA]) log Ka = pKa pKa = log ([H+]*[A]/[HA]) Now refer to the log rules above, to rewrite the right hand side: pKa = log[A]  log[H+]  (log[HA]) You know that pH = log [H+], so substitute in the above: pKa = log[A] +pH +log[HA] Rearrange so that pH is the subject of the equation: pH = pKa + log [A]  log [HA] Log rules state that the expression (log[HA]) means divide by [HA] Rewrite as: pH = pKa + log ([A]/[HA]) That is how to derive the HH equation.
From Youtube
Buffer Solutions :General Chemistry II lecture covering buffer solutions of acids and their conjugate bases. Buffer solutions are designed to maintain the pH of solution by reacting with small amounts of added acid or base. We show how to calculate the pH of a buffer solution using the HendersonHasselbalch equation, and how to select an appropriate acidbase combination to prepare a buffer solution at any desired pH.
1. Buffer solutions  pH of a buffer :Calculation of pH of a buffer solution in which the concentration of weak acid and conjugate base are the same. First of a three part question about buffer solutions. Suitable for Alevel, International Baccalaureate or equivalent. Content includes pH calculations using Ka, change in pH on addition of a small amount of H+, and comparison to adding the same amount of H+ to water.