Three liquids are given to you. One is hydrochloric acid, another is sodium hydroxide and third is a sugar solution. How will you identify them? You have only turmeric indicator. Write in details
Answers
Answer:
✏ Answer
\begin{gathered}(x + 3y) {}^{3} = x {}^{3} + 27y {}^{3} + 3(xy3)(x + 3y) \\ (6) {}^{3} = x {}^{3} + 27y { }^{2} + 3 \times 3(2)(6) \\ 216 = {x}^{3} + 27y {}^{3} + 108 \\ 216 - 108 = x {}^{3} + 27y {}^{3} \\ x {}^{3} + 27y {}^{3} = 108\end{gathered}
(x+3y)
3
=x
3
+27y
3
+3(xy3)(x+3y)
(6)
3
=x
3
+27y
2
+3×3(2)(6)
216=x
3
+27y
3
+108
216−108=x
3
+27y
3
x
3
+27y
3
=108
When we place turmeric solution in contact with bases it turns into red colour and generally It is not affected by acids and neutral substances.
Now, we can distinguish between hydrochloric acid, sodium hydroxide and a sugar solution by applying turmeric solution to three test tubes, we will observe that
(1) test tube A (Hydrochloric acid): No change in colour will take place because turmeric is not affected by acids.
(2) test tube B (sodium hydroxide): Turmeric solution will change its colour to red because turmeric solution changes its colour to red when we place it in contact with bases.
(3) test tube c (sugar solution): No change in
color will take place because sugar solution is neither basic nor acidic in nature
Therefore, we are only able to identify basic solution using turmeric as an indicator