Question

In: Chemistry

Methyl orange changes from red at pH 3.1 to yellow at pH 4.4. Would methyl orange...

Methyl orange changes from red at pH 3.1 to yellow at pH 4.4. Would methyl orange be a suitable indicator for the titration of propanoic acid (pKa=4.87, Ka=1.34x10^-5) with 0.01 M NaOH? Explain. Phenolphthalein changes from colorless at pH 8.0 to pink at pH 9.6. Would phenolphthalein be a suitable indicator for the titration of propanoic acid with 0.01 M NaOH? Explain.

Solutions

Expert Solution

Indicator- is a dye(substance) which changes colour over a short pH range.

an indicator is usually weak acid or weak base for acid-base titration.the ccolour change occurs when a proton transferred to form a conjugate base or a conjugate acids of a different colour

Methyl orange (MeOH) is represents as

MeOH Me+ + OH-

(Yellow in base) (Red in acid)

When a strong acid is added,equilibrium of MeOH shifts towards the right and the colour change Red is detected.

Phenolphthalein (PhH)

PhH    Ph+ + H+

(Colourless) (pink in base)

PhH is a colourless in acid solution and pink in basic solution. When a strong base such as NaOH is added it displaces the equilibrium towards right direction and there is a colur change.

As in given question titration of propanoic acid(weak acid as organic acid mostly weak) and NaOH(strong base),the pH range near the end point is 8-10.since the pH lies on alkaline side.phenolphthalein (pH range 8-9.6 or 9.8) and thymol blue(pH range 8.6-9.6)are suitable indicators for such titrations.methyl orange (pH range 3.1-4.4)cannot be used for such titrations.also propanoic acid is weak acid does not furnish sufficient number of H+ ions.these H+ ions is not sufficient to combine with sufficient number of OH- ions of methyl orange to form water and to shift the equlibrium in favour of red colouration Me+ ions.

MeOH Me+ + OH-

CH3CH2COOH   CH3CH2COO- + H +

CH3CH2COO- + Me+ CH3CH2COOMe

H+ + OH-    H2O

However,the salt of CH3CH2COOMe does not furnish sufficient red ions because of hydrolysis

CH3CH2COOMe + H2O CH3CH2COOH + MeOH


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