
Indicator $ {\text{pH range}} $ Methyl violet $ 0.3 - 1.8 $ Methyl orange $ 2.8 - 3.8 $ Congo red $ 2.8 - 4.8 $ Methyl red $ 3.8 - 6.1 $ Bromothymol blue $ 6.0 - 7.9 $ Phenol red $ 6.8 - 8.6 $
The table above shows the $ {\text{pH}} $ ranges of some common indicators. Given that $ {K_a} $ is $ 7.3 \times {10^{ - 10}} $ for boric acid $ ({H_3}B{O_3}) $ , Choose an indicator that can be used for the titration of $ 0.10M{\text{ }}K{H_2}B{O_3} $ with $ 0.10M{\text{ }}HCl $
(A) Methyl orange
(B) Congo Red
(C) Methyl Red
(D) Phenol Red
| Indicator | $ {\text{pH range}} $ |
| Methyl violet | $ 0.3 - 1.8 $ |
| Methyl orange | $ 2.8 - 3.8 $ |
| Congo red | $ 2.8 - 4.8 $ |
| Methyl red | $ 3.8 - 6.1 $ |
| Bromothymol blue | $ 6.0 - 7.9 $ |
| Phenol red | $ 6.8 - 8.6 $ |
Answer
493.8k+ views
Hint :To solve this, we have to write down the chemical reaction of $ K{H_2}B{O_3} $ with $ HCl $ and the products formed. Then find concentration of products at the equivalence point with the help of $ {K_a} $ value given in the question. With the help of concentration of the hydronium ion, find the $ pH $ . Then check in which indicator range the $ pH $ lies.
Complete Step By Step Answer:
The reaction of $ K{H_2}B{O_3} $ with $ HCl $ is as follows:
$ K{H_2}B{O_3} + HCl \to {H_3}B{O_3} + KCl $
At equivalence point the concentration of $ {H_3}B{O_3} $ produced $ = 0.050M{\text{ }}{H_3}B{O_3} $ then ionizes so:
$ {H_3}B{O_3} + {H_2}O \to {H_3}B{O_3} + {H_3}{O^ + } $
$ 0.05M $ $ x $ $ x $
$ {k_a} = \dfrac{{[{H_3}{O^ + }][{H_3}B{O_3}]}}{{[{H_3}B{O_3}]}} = \dfrac{{{x^2}}}{{0.05}} $
i.e. $ 7.3 \times {10^{ - 10}} = \dfrac{{{x^2}}}{{0.05}} $
$ x = 6.04 \times {10^6} $
Using the concentration of hydronium ion, we can find the $ pH $ i.e.
$ pH = - \log (6.04 \times {10^6}) = 5.22 $
This $ pH $ is in the range of methyl red indicators which will be suitable for the given titration.
Therefore, the correct option is C.
Note :
The substances that change color when they are added to alkaline or acidic solutions are called indicators. Methyl orange is a $ pH $ indicator used in titrations due to its clear and distinct color variance at different values of $ pH $ . It shows red color in acidic medium and yellow color in basic medium. As it changes its color at the $ p{K_a} $ of a mid-strength acid, it is used in titration of acids. Congo red serves as an acid-base indicator as it turns red in the presence of alkalies and blue in the presence of acids.
Complete Step By Step Answer:
The reaction of $ K{H_2}B{O_3} $ with $ HCl $ is as follows:
$ K{H_2}B{O_3} + HCl \to {H_3}B{O_3} + KCl $
At equivalence point the concentration of $ {H_3}B{O_3} $ produced $ = 0.050M{\text{ }}{H_3}B{O_3} $ then ionizes so:
$ {H_3}B{O_3} + {H_2}O \to {H_3}B{O_3} + {H_3}{O^ + } $
$ 0.05M $ $ x $ $ x $
$ {k_a} = \dfrac{{[{H_3}{O^ + }][{H_3}B{O_3}]}}{{[{H_3}B{O_3}]}} = \dfrac{{{x^2}}}{{0.05}} $
i.e. $ 7.3 \times {10^{ - 10}} = \dfrac{{{x^2}}}{{0.05}} $
$ x = 6.04 \times {10^6} $
Using the concentration of hydronium ion, we can find the $ pH $ i.e.
$ pH = - \log (6.04 \times {10^6}) = 5.22 $
This $ pH $ is in the range of methyl red indicators which will be suitable for the given titration.
Therefore, the correct option is C.
Note :
The substances that change color when they are added to alkaline or acidic solutions are called indicators. Methyl orange is a $ pH $ indicator used in titrations due to its clear and distinct color variance at different values of $ pH $ . It shows red color in acidic medium and yellow color in basic medium. As it changes its color at the $ p{K_a} $ of a mid-strength acid, it is used in titration of acids. Congo red serves as an acid-base indicator as it turns red in the presence of alkalies and blue in the presence of acids.
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