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12.0 14.0Colorless - Orange1,3,5-trinitrobenzena10.1 12.0Yellow VioletAlizarin yellow R

9.3 10.6Colorless BlueTymolphthlaein

8.0 9.6Colorless PinkPhenolphthalein

7.0 9.0Yellow Bluep- -naphtolphthalein

6.8 8.4Yellow RedPhenol red

6.8 -8.0Red YellowNeutral red

6.0 7.6Yellow BlueBromcresol blue

5.2 6.8Yellow PurpleBroncresol purple

5.0 7.0Colorless - Yellowp-nitrophenol

4.8 5.4Purple GreenMethyl purple

4.5 8.3Red BlueLitmus4.2 6.2Yellow RedMethyl red

3.8 5.4Yellow BlueBromcresol green

3.1. 4.4.Red YellowMethyl orange

3.0 4.6Yellow BlueBromphenol blue

2.9 4.0Red YellowYellow methyl

2.0 4.0Colorless Yellow2,6-dinitrophenol

1.2 2.8Red - YellowThymol Blue

0.1 0.8Colorless - YellowPicric acid

pH rangeColor change with pH increasedIndicator

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pH(Contd)

Solution SBI 2008 34

As we can see from the table above, eachindicator has its specific range.

So when we want to do titration where theprediction of pH will be around 8.0 9.0,the indicator used will be

Phenolphthalein

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pH(Contd)

Solution SBI 2008 35

Now, lets back to the pH calculation. We havelearned how to calculate pH for strong acid andbase . Can the same formulae be used straightway for weak acid and base?

Well, the formulae to calculate pH and pOH arebasically the same, but how to calculate [H + ]and [OH -] are different.

As an example how to calculate concentration of [H + ] lets see what happen for a weak acid HA

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pH(Contd)

Weak acid HA (with the valency

of 1) will be ionized as follows:H+ A-+HA

Solution SBI 2008 36

The Ka value for this acid is then

K a =[H

+][A

-]

[HA]

As the concentration acid dissociated istoo small compared to the initial con-centration, so [HA] = [Acid] x = [Acid]

With x the acid which has been dissociatedThe equilibrium equation is then:In this case:

[H + ] = [A -], so[H+][A-] = [H+] 2

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pH(Contd)

Solution SBI 2008 37

or

Ka [acid]

K a =[H+]

[HA]

2K a =

[H+]

[Acid]

2

[H+] =

So we calculate [H +] as

When ionization degree ( )is known, then tocalculate the acid which has been dissociated as:

[H + ] = x [acid]

To find the concentration of [OH -] from weak base, thesame way as the weak acid applies .

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pH(Contd)

Solution SBI 2008 38

Example 1.Please calculate pH of 6.0 g CH 3COOH which is

dissolved in 1000 mL solution (assume Ka as 1x 10 -5 )

Answer: Calculate the [H+

]First find the concentration of acid.M = (mass/Mr) x (1000/volume) =

(6/60)(1000/1000)

= 0.1 M

[H+

] = V1 x 10-5

x 0.1= -3

Ka [acid][H+

] =

pH = -log [H +]= -log 10 -3

= 3

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pH(Contd)

Solution SBI 2008 39

Example 2. 1.75 g of NH 4OH (Mr= 35 and Kb 1 x10 -5) is dissolved up to 500 mL, calculate the pHof this solution!

Question of Solution1. Which of the following solution has the highestmolarity in 1000 mL solution? (Mr H 2SO 4 =98,HCl=36.5, HBr=81, H 3PO 4=98)

a. 1.96 g H 2SO 4 b. 1.46 g HCl c. 0.81 g HBrd. 0.98 g H 3PO 4

2. Based on question no. 1, the pH of H 2SO 4 will

be- -

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pH(Contd)

Solution SBI 2008 40

Question of Solution3. Which of the following base solution can

neutralize HCl solution in question 1?a. 40 mL NaOH 0.5 M b. 20 mL NaOH 1 M c.40 mL NaOH d. 60 mL NaOH 0.5 M

4. Based on question no. 1, if of H 2SO 4 is 1, thesolution will boil at (=1, Kb of water=0.52C/mole)

a. 100.0312 C b. 100.0624 C c. 100.0156 C d. none is correct

5. If R=0.082 Latm/K.mole and temperature is27 C, H3PO 4 is 0.85, the osmotic pressure of H3PO 4 in question 1 will be.. (in atm)

a. 0.123 b. 0.246 c. 0.492 d. 0.984

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pH(Contd)

Solution SBI 2008 41

Question of Solution6. Which of the following HCl has a concentration

equal to HBr in question 1 (all HCl in 100 mL,Mr 36.5)?a. 36.5 g b. 3.65 g c. 3.65 mg d. 36.5

mg7. What is ionization degree of HCN 1 M if

Ka 6.25 x 10 -10

a. 6.25 x 10 -5 b. 6.25 x 10 -2 c. 2.5 x 10 -5 d. 2.5

x 10 -28. pH of HA acid which concentration of 0.08 M is

3. Calculate the Ka and !!!! (essay)