Question

# a) An unknown indicator (called “HIn”) has a Ka of 4.0 x 10-6. The color of...

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

b)

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

c)

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]

= 2.3010

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