ionic equilibria
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ammonia is a weak bronsted lowry base of Kb value of 1.78x 10^-5 moldm-3
a 25.0cm3 sample of aqueous ammmonia was placed in a conical flask, when titrated against the 0.200moldm-3 hydrochloric acid, 17.50cm3 of acid was required for complete neutralization.
calculate the initial concentration of aqueous ammonia in the conical flask.
*for this question can i use the stoichiometric ratio that no. of moles of NH3 = no. of moles of HCl? then calculate the initial conc?
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another question,how to find the concentration of ethanoic acid from the graph of pH against volume of NaOH added?
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Originally posted by chocolates-xed:
ammonia is a weak bronsted lowry base of Kb value of 1.78x 10^-5 moldm-3
a 25.0cm3 sample of aqueous ammmonia was placed in a conical flask, when titrated against the 0.200moldm-3 hydrochloric acid, 17.50cm3 of acid was required for complete neutralization.
calculate the initial concentration of aqueous ammonia in the conical flask.
*for this question can i use the stoichiometric ratio that no. of moles of NH3 = no. of moles of HCl? then calculate the initial conc?
Just use M1V1 = M2V2 where M is molarity and V is vol. Just becareful of your units.
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Originally posted by chocolates-xed:
another question,how to find the concentration of ethanoic acid from the graph of pH against volume of NaOH added?
From the graph, u can get the vol of NaOH added for neutralisation, i believe u should have their concentration (molarity) so use the above method to calcuate the concentration.
Since this is between weak acid and strong base, neutralisation should be around pH 9-10 if i'm not wrong. Just look at your graph.
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have a few more questions..
1) ammonia is a weak bronsted lowry base of Kb value of 1.78x 10^-5 moldm-3
a 25.0cm3 sample of aqueous ammmonia was placed in a conical flask, when titrated against the 0.200moldm-3 hydrochloric acid, 17.50cm3 of acid was required for complete neutralization.
how to calculate the pH at the equivalence pt of the titration?
2) Solution X contains 0.100moldm-3 calcium chloride and 0.100 moldm-3 barium chloride. Predict and explain, with the aid of relevant calculations, what would happen when equal volumes of solution x and 0.00500moldm-3 sodium floride are mixed together
3) State and explain how the solubility of barium floride would change if it is dissolved in an aqueous solution of KF
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Originally posted by chocolates-xed:
have a few more questions..
1) ammonia is a weak bronsted lowry base of Kb value of 1.78x 10^-5 moldm-3
a 25.0cm3 sample of aqueous ammmonia was placed in a conical flask, when titrated against the 0.200moldm-3 hydrochloric acid, 17.50cm3 of acid was required for complete neutralization.
how to calculate the pH at the equivalence pt of the titration?
2) Solution X contains 0.100moldm-3 calcium chloride and 0.100 moldm-3 barium chloride. Predict and explain, with the aid of relevant calculations, what would happen when equal volumes of solution x and 0.00500moldm-3 sodium floride are mixed together
3) State and explain how the solubility of barium floride would change if it is dissolved in an aqueous solution of KF
Tips :
Q1) Do ICF (Initial Change Final) and ICE (Initial Change Equilibrium), then use Ka expresion (for proton dissociation from the conjugate acid ammonium cation) to find [H+] and thus pH.
Q2) The question lacks Ksp data. But predictably, calcium fluoride will be selectively precipitated over barium fluoride. Support your answer with calculations involving ionic and solubility products for both ionic compounds.
Q3) Explani using Le Chatelier's principle and common ion effect. Molar/Mass solubility of barium fluoride will decrease.
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hmmm for qn 1 right,
can i use this method?
NH3 + H20 <-> NH4+ OH-
use the ice table, i get [OH-] = 1.57 x 10^-3
then find the pOH , then use pH = 14 - pOH
but its not really the answer i was looking for, cause the next subpart of the question asks which type of indicator to use and they provided indicators from pH 0.0 to 9.6
but my pH is like 11.2
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Originally posted by chocolates-xed:
hmmm for qn 1 right,
can i use this method?
NH3 + H20 <-> NH4+ OH-
use the ice table, i get [OH-] = 1.57 x 10^-3
then find the pOH , then use pH = 14 - pOH
but its not really the answer i was looking for, cause the next subpart of the question asks which type of indicator to use and they provided indicators from pH 0.0 to 9.6
but my pH is like 11.2
At equivalence point, you only have the protonated conjugate acid, not the deprotonated conjugate base. So your equation for your ICE table is NH4+ ---> H+ + NH3 or alternatively, NH4+ + H2O ---> H3O+ + NH3
The correct pH at equivalence point would be less than 7, due to protons dissociated from the conjugate acid.