For each of the following reactions A and B,
(i) identify the 2 acids nad 2 bases present
(ii) suggest, with reasons, which ion or molecule i fthe stronger acid, and which is the stronger base.
A NH3 + H2O <----> NH4+ + OH- Kc =1.8 X 10^-5 mol/dm3
B C6H5O- + CH3CO2H <---> C6H5OH + CH3CO2 Kc=1.3 X 10^5 mol/dm3
Need help with (ii). I think we have to identify the stronger acid and base in EACH equation.
For A,
NH3 is a base, OH- is the conjugate base. H2O is the acid and NH4+ is the conjugate acid.
Since NH3 is a weak base, NH4+ forms a STRONG conjugate acid relative to H2O. (from my own understanding, pls correct where necessary) How do I find who is the stronger acid?
Also need help with reaction B.
Thank you.
Have one more question.
Use the relationshop
pH = pKa - lg([acid]/[salt]) tp calculate the pH of a solution which is 0.40 moldm-3 w.r.t ethanoic acid and 0.20mol dm-3 w/r/tsodium ethanoate
[take Ka for ethanoic acid as 1.80 X 10^-5 mol/dm3)
(ans 4.44)
Calculate the change of pH of 1.0 moldm-3 in (a) when 0.050 mol of solid NaOH is added (assume no change in vol) (ans 0.155)
I am not sure what the 1.0mol dm-3 is there for... well isnt it stated that there were 0.40moldm-3 of aci and 0.20 moldm-3 of salt initially? How do I make use of what I found in part (a) to do this part?
What is the change of pH when 0.050mol of NaOH is aded to 1.0mol dm-3 of water (5.7 pH units)
I only need help with the BOLD parts. thanks.
Originally posted by Audi:For each of the following reactions A and B,
(i) identify the 2 acids nad 2 bases present
(ii) suggest, with reasons, which ion or molecule i fthe stronger acid, and which is the stronger base.
A NH3 + H2O <----> NH4+ + OH- Kc =1.8 X 10^-5 mol/dm3
B C6H5O- + CH3CO2H <---> C6H5OH + CH3CO2 Kc=1.3 X 10^5 mol/dm3
Need help with (ii). I think we have to identify the stronger acid and base in EACH equation.
For A, NH3 is the base, H2O is the acid, NH4+ is NH3's conjugate acid, OH- is H2O's conjugate base. Since position of equilibrium lies to the left, OH- ion is a stronger base than NH3, and NH4+ is a stronger acid than H2O.
For B, phenoxide ion is the base, ethanoic acid is the acid, phenol is phenoxide ion's conjugate acid, and ethanoate ion is ethanoic acid's conjugate base. Since position of equilibrium lies to the right, phenoxide ion is a stronger base than ethanoate ion, and ethanoic acid is a stronger acid than phenol.
As an exercise in Organic Chem, you should also be able to come to the same conclusion for which are the stronger acids & bases, by looking at electronegativities, induction and resonance considerations of the charges (ie. analysis of electronics) of LHS vs RHS species, as an alternative to using Kc values. If you do this correctly, the conclusions should concur with the Kc values.
Originally posted by Audi:Have one more question.
Use the relationshop
pH = pKa - lg([acid]/[salt]) tp calculate the pH of a solution which is 0.40 moldm-3 w.r.t ethanoic acid and 0.20mol dm-3 w/r/tsodium ethanoate
[take Ka for ethanoic acid as 1.80 X 10^-5 mol/dm3)
(ans 4.44)
Calculate the change of pH of 1.0 moldm-3 in (a) when 0.050 mol of solid NaOH is added (assume no change in vol) (ans 0.155)
I am not sure what the 1.0mol dm-3 is there for... well isnt it stated that there were 0.40moldm-3 of aci and 0.20 moldm-3 of salt initially? How do I make use of what I found in part (a) to do this part?
What is the change of pH when 0.050mol of NaOH is aded to 1.0mol dm-3 of water (5.7 pH units)
I only need help with the BOLD parts. thanks.
I recommend you do not use the Henderson-Hasselbach equation. Instead just use the formulaic definition of Ka, which (like any Kc) is basically the molarities of RHS over the molarities of the LHS, for the proton dissociation equation of the weak acid.
Do the ICE table for CH3COOH <---> H+ + CH3COO-. You'll notice that for a buffer system, the equilibrium molarities of the weak acid and its conjugate base are approximated back to initial molarities, which means you can skip the ICE table if you wish. Your ICE table should have units as molarities.
Sub the equilibrium molarities into the Ka formula, make molarity of protons [H+] the subject, and hence find pH. (4.44)
Do the ICF table for adding the sodium hydroxide. Usually we use moles as units for the ICF table, but since the volume of solution does not change in this instance, we can work directly in molarities.
Your ICF table for CH3COOH + OH- --> CH3COO- should have initial molarities 0.4, 0.05 and 0.2. Your final (not 'equilibrium') molarities should be 0.35, 0, 0.25.
Plugging these values into your Ka formula (for acidic buffers, you can use Ka of the acid or Kb of the conjugate base), and making molarity of protons the subject, you obtain pH = 4.5986 or 4.6 to 3 sf. Hence the change in pH is approx 0.1586 or 0.159 to 3 sf. (don't automatically assume your school handout's stated answer is always correct).
As to the adding of NaOH(s) to pure water (pH 7 at rtp), notice that NaOH is a strong alkali. Hence, no Kb value is required. The pOH is 1.3 hence pH is 12.7 hence change in pH is 5.7.