Question

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. (K_a 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 K_a 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.

Answer 1

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