Question

Identify the indicator that can be used at the highest pH. O alizarin yellow R phenolphthalein alizarin methyl red bromophenol blue Submit Request Answer

Answer 1

1-

A pH indicator is a compound which changes its color within a short pH range. Each indicator has its own pH range. Beyond that it doesnt change color so become uselsss afetr its pH range.

Now among the given options, the highest pH range is of Alizarin yellow R (pH = 10 - 12). At lower pH i.e 10, it is yellow in color and as it reaches to pH 12, it changes to red color.

Beloe pH 10 and above pH 12, Alizarin yellow R is ineffective.

2-

Here the solution is a buffer and the pH of buffer is calculated by Henderson- Hasselblach equation-

pH = pKa + log [salt]/ [weak acid]

where

pKa = -log Ka

In the question we have a weak acid Acetic acid (Ka = 1.8 * 10-5). Now when we add a strong base like KOH to this,s the recation takes place as-

CH3CO2H (weak acicd) + KOH --------> CH3CO2K (salt) + H2O

Now given

volume of acid takenn (V1)= 25 mL

volume of base added (V1)= 25 mL

initial concentration of acid (M1) = 0.1 M

initial concentration of base added (M1) = 0.1 M

After addition of both, new volume (V2) = 25 + 25 = 50 mL

So

new concentration of acid (M2) = M1V1 / V2 = 0.1 M * 25 mL / 50 mL = 0.05‬ M

new concentration of base (M2) = M1V1 / V2 = 0.1 M * 25 mL / 50 mL = 0.05‬ M

So final concentrations will be-

Reaction	CH3CO2H (weak acicd) +	KOH -------->	CH3CO2K (salt)
Initial	0.05‬ M	0.05‬ M	0
Change	- 0.05‬ M	- 0.05‬ M	+ 0.05‬ M
Equilibrium	0	0	0.05‬ M

That mens after the reaction, we have only  0.05‬ M of CH3CO2K left. So at this pH of the solution is due to 0.05‬ M of CH3CO2K

Now we know CH3CO2K dissociate in water as-

CH3CO2K ----------> CH3CO2- + K+  

Now we know CH3CO2- is the conjugate base of CH3CO2H, which reacts in water as-

CH3CO2- + H2O --------------> CH3CO2H + OH-

where

Kb = Kw / Ka

= 10-14 / (1.8 * 10-5)

= (0.55 * 10-5)

Again we know

Kb = [CH3CO2H] * [OH-] / [CH3CO2-]

to find these concentrations lets form the ICE table as-

Reaction	CH3CO2- ------->	CH3CO2H +	OH-
Initial	0.05‬ M	0	0
Change	- x	+x	+ x
Equilibrium	0.05‬ - x M	x	x

putting the values-

Kb = [CH3CO2H] * [OH-] / [CH3CO2-]

(0.55 * 10-5) = [x] * [x] / [0.05‬ - x ]

(0.55 * 10-5) * [0.05‬ - x ] = x2  

(0.0277 * 10-5) - (0.55 * 10-5)x = x2  

x2 + (0.55 * 10-5)x - (0.0277 * 10-5) = 0

Solving this-

x = 0.000523 M

So

[OH-] = x = 0.000523 M

So

pOH = -log OH = -log 0.000523 = 3.28

So

pH = 14 - pOH = 14 - 3.28 = 10.72‬