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a) An unknown indicator (called “HIn”) has a Ka of 4.0 x
10-6. The color of the neutral form is green, and that
of ionized form is red. The indicator is added to a HCl solution,
which is then titrated against a NaOH solution. At what pH will the
indicator change color?
b) 350. mL of a NaOH solution was added to 500. mL of 2.50 M HNO2. The pH of the mixed solution was 1.75 units greater than that of the original acid solution. Calculate the molarity (in M) of the initial NaOH solution. (Ka of HNO2 is 4.5x10-4)
c) 30.0 mL of 0.100M HCl is titrated with a 0.100 M CH3NH2 solution. Calculate the pH values of the solution at the following titration points (the Ka of CH3NH3+ is noted to be 2.0x10-11): After 30.0 mL of the CH3NH2 solution has been added to the initial acid volume.
a) pKIn of indicator = -log 4.0x10-6 = 5.3979
The reaction is
HIN + OH- ----------------> In- + H2O
The pH of this buffer is
pH = pKIn + log [IN-]/[HIn]
If the ratio [In-][HIN] > 10, the ionisation is almost complete and color changes.
Thus the pH at which color changes = pKIn + 1 = 5.3979+1 = 6.3979
molarity of acid = 2.50 M
pka = -log 4.5x10-4 = 3.347
The pH of acid solution = 1/2[pKa -log C] = 1/2[ 3.347 -log2.50] =2.949
Thus the pH of buffer addition of NaOH = 2.949 + 1.75= 4.699
NaOH + HNO2 ----------------> NaNO2 + H2O
350xa 500x2.50=1250 --- initial mmoles
0 1250-350a 350a --- at equilibrium
the pH of this buffer is given by Hendersen equation as
pH = pKa + log [conjugate base]/[acid]
4.699= 3.347 + log [conjugate base]/[acid]
So [conjugate base]/[acid] =22.49 = [350a]/ [1250 -350a]
solving for a , a= 3.419
HCl + CH3NH2 ------------------------> CH3NH3Cl
30x0.1=3 30x0.1=3 0 inital mmoles
0 0 3
[salt ] formed = mmol/volume = 3/60 = 0.05 M
pKb of CH3NH2 = -log 2.0x10-11 = 10.6989
pH of the solution is given as
pH = 1/2[ pKw -pKb - logC] = 1/2[ 14- 10.6989 -log 0.05]
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