A 0.20 M solution of sodium nitrile, NaNO_2:. has a pH of 8.57. Write the chemic
ID: 938770 • Letter: A
Question
A 0.20 M solution of sodium nitrile, NaNO_2:. has a pH of 8.57. Write the chemical equation showing why this salt has the given Calculate K_1,for the anion and K_a for the corresponding conjugated acid given the measured pH, Draw the Lewis structure for momochlroacetic acid. The pk_a of monochlormactic acid is 2.865 at 25^degree . Compare this pK_a to that of acetic acid and sugegest a explanation for the large differences in pk_a values. Estimate the pH of a 0.10 M tuonositlcciaccuc acid. What is the approximate percent of ionization Results must be reported to the correct number of significant figures. At 60 ^degree C, Kw = 9.5 Times 10 ^-14. What is the pH of pure water at this temperature? Why can we say that this water is neither acidic nor basic? At 25 ^degree C, the water in a natural pool of water in one of the western states was found to contain hydroxide ions at a concentration of 4.7 Times 10^-7 g OH per liter, Calculate the pH of this water State if it is acidic, basic, or neutral. The first antiseptic to be used in surgical operating rooms was phenol, C_6H_5OH, a weak acid and a potent bactericide. A 0.550 M solution of phenol in water was found to have a pH of 5.07. Write the chemical equation for the equilibrium involving C_6H_5OH in the solution. Write the equilibrium law corresponding to K for C_6H_5OH. for this acid?Explanation / Answer
1. a.
NaNO2 ---> Na+ + NO2- --- dissociation
Chemical equation : NO2- + H2O <==> HNO2 + OH- ---reacts with H2O and acts as base
Kb = [HNO2][OH-]/[NO2-]
b. pH = 8.57
pOH = 14 - 5.43
[OH-] = 3.71 x 10^-6 M
Ka = 1 x 10^-14/[(3.71 x 10^-6)^2/0.2] = 1.45 x 10^-4 for conjugate base
Ka = (3.71 x 10^-6)^2/0.2 = 6.89 x 10^-11 for anion
2. a. Given below is the Lewis structure for monoachloroacetic acid.
b. Chlorine is electronegative atom and pulls the shared pair of electrons of Cl-C bond towards it by inductive effect. The carbon thus pulls electrons towards itself from the O-H bond, thus making the H more electropositive in nature and easy to loose. This effect is not present in acetic acid and thus monochloroacetic acid is more acidic than acetic acid.
c. pH of 0.1 M monochloroacetic acid
Ka = x^2/0.1 = 1.8 x 10^-5
x = [H+] = 1.34 x 10^-3 M
pH = 2.87
percent isonization = (1.34 x 10^-3/0.1) x 100 = 1.34%
Connecting to lab questions
1) Kw = 9.5 x 10^-14
pH = -log(9.5 x 10^-14/2) = 13.32
The water contains equal amount of [H+] and [OH-] ions. The Kw is very ow, such that minimum ionisation of water exists at this temperature, which makes it neither acidic not basic.
2) a. pH = -log(Kw/Koh) = -log(1 x 10^-14/4.7 x 10^-7) = 7.67
b. The solution is slightly basic in nature.
3) Equation : C6H5OH <==> C6H5O- + H+
b. Ka = [C6H5O-][H+]/[C6H5OH]
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