Question

pH Measurements – Buffers and Their Properties 1. A solution of a weak acid was tested with the indicators used in this experiment. The colors observed were as follows: Methyl violet Thymol blue Methyl yellow

What is the approximate pH of the solution?_______________ violet Congo red orange-red yellow Bromcresol green green yellow

2. A solution of Na2CO3 has a pH of 10. The CO32 ion is the conjugate base of the HCO3  ion. Write the net ionic equation for the reaction which makes a solution of Na2CO3 basic. (See Equation 8.11)

3. The pH of a 0.10 M HCN solution is 5.20. a. What is [H3O+] in that solution? _______________ M b. What is [CN ]? What is [HCN]? (Where do the H+ and CN  ions come from?) [CN ] = _______________ M; [HCN] = _____________ M c. What is the value of Ka for HCN? What is the value of pKa? Ka = _______________ pKa = _______________ 8.10

4. Formic acid, HFor, has a Ka value of 1.8 × 10 4. A student is asked to prepare a buffer having a pH of 3.40 from a solution of formic acid and a solution of sodium formate having the same molarity. How many milliliters of the NaFor solution should she add to 20 mL of the HFor solution to make the buffer? (See discussion of buffers.) _______________ mL

5. How many mL of 0.10 M NaOH should the student add to 20 mL 0.10 M HFor if she wished to prepare a buffer with a pH of 3.40, the same as in Problem 4? _______________ mL

Answer 1

1. According to the color change approximate pH = 5

2.

CO₃^2- is the conjugate base of HCO₃^- ion.

Na₂CO₃ + H₂O   H₂CO₃ + 2 NaOH

ionic equation: 2 Na⁺ + CO₃^2- + H₂O   H₂CO₃ + 2 Na⁺ + 2OH^-

Net ionic equation; CO₃^2- + H₂O   H₂CO₃ + 2OH^-

3.

pH = - log [H₃O⁺] = 5.20

or, [H₃O⁺] = 10^-5.20 = 6.31 *10^-6 M

HCN + H₂O H₃O⁺ +CN^-

hence, [H₃O⁺] = [CN^-] = 6.31*10^-6 M

given, [HCN] = 0.10 M

now, for weak acid HCN

pH = (pKa -log 0.10)

5.20 = ( pKa - log 0.10)

or, pKa  - log0.1 = 10.4

or, pKa +1 = 10.4

or, pKa = 9.4

or , Ka = 3.98*10^-10

4.

Henderson-Hasselbalch equation is

pH = pKa + log

or, 3.40 = - log(1.8*10^-4) + log

or, 3.40 = 3.745 + log

or, log = - 0.075

or,   = 10^-0.075 = 0.8414

given , initial molarity of NaFor and HFor are same

hence, volume of NaFor added = Volume of HFor * 0.8414 = 20* 0.8414 = 16.828 mL

5.

or, 3.40 = 3.745 + log

or, = 0.8414

given,

initial mmoles of HFor = 20*0.10 = 2

let x mmoles NaOH added

	HFor	NaOH	NaFor
before mmoles	2	x	0
change in mmoles	-x	-x	+x
after mmoles	2-x	0	x

now,

= 0.8414

or, (x/2-x) = 0.8414

or, x = 1.6828 - 0.8414 x

or, 1.8414 x = 1.6828

or, x = (1.6828/1.8414 ) = 0.9138

hence mmoles of NaOH added = 0.9138

or, mmoles= molarity * volume(mL)

or, Volume = (mmoles/molarity) = (0.9138/0.1) = 9.138 mL