ammonia buffer solution
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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: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: 1.8E-5 = [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: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: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 H-H 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 H-H equation.