Shown is an acid (HA) titrated by NaOH. The black dot shows the equivalence poin
ID: 693874 • Letter: S
Question
Shown is an acid (HA) titrated by NaOH. The black dot shows the equivalence point. Is this a weak or strong acid? What are the principle species in the region indicated by the arrow? 1) 14 12 10 A) Weak acid; HA B) Weak acid; HA, A C) Strong acid: HO D) Strong acid; HA, A 6 2 0.00 20nP NaOH 40.00 2) In which of the following aqueous solutions would you expect AgBr to be the most soluble? A) 0.1 M HBr B) 0.1 M AgNOs C) 0.1 M NaBr D) Pure water 3) Triethylamine, (CH CH2)N, has a Kb of 5.2 x104 at 25°C. What is the neutralization equilibrium constant (Ka) if it is reacted with HI? A) 5.2 × 10-18 B) 1.9 x 10-11 C) 1.9 x 103 D) 5.2 x 1010 4) You titrate 1.0 L of 0.10 M HC3Hs02 (monoprotic acid) halfway to its equivalence point using 1.0 M NaOH. What is the pH? (K. for HC HsO2 is 1.3 x10) A) 1.00 B) 3.09 C) 4.89 D) 8.92 Useful Information: pH =-log[H3O+] ; K, = [H30+][AT] ; Kb = [OH-][BH+], Kn Kw (strong acid-strong base) Acid] = Ka/Kw (weak acid-strong base) = Kb/Kw (strong acid-weak base) KaKbKw (weak acid-weak base) Strong acids: HI, HCI, HBr, HCIO4, HCIOs, H2SO4, HNO Strong bases: any salts of group 1 metals and hydroxide ionExplanation / Answer
1) The pH at the equivlanece point is 9.0 approximately, that is >7 . that shows the acid is weak , as the salt formed is basic. Thus the option is A [weak acid]
2)OPtion D
AgBr is most soluble in pure water as the other solutions have common ion.
In the presence of common ion ,the solubility of sparingly soluble salt further decreases.
3) OPtion D
Kn = Kb/Kw as the reaction is between strong acid and weak base
Kn = 5.2x10-4 /1.0x10-14
= 5.2 x 1010
4)It is the titration of weak aacid with strong base.
HA + OH- ------------> A- + H2o
At half equivalence [HA] = [A-]
and the solution behaves as a buffer of weak acid and its conjugate base.
pKa =5- log 1.3 =4.886
The pH of suc buffer is given by Hendersen equation as
pH = pKa + log [[conjugate base]/[acid]
= 4.886 + log (1/2)/(1/2)
= 4.886
At the half equivalence point pH = pKa
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