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Boric acid can be titrated against sodiu...

Boric acid can be titrated against sodium hydroxide using methyl orange as indicator only in the presence of polyhydroxy compounds wlike catechol, manniton etc. explain.

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### Step-by-Step Solution: 1. **Understanding Boric Acid and its Reaction**: - Boric acid (H₃BO₃) can react with sodium hydroxide (NaOH) to form sodium metaborate (NaBO₂) and water. The reaction can be represented as: \[ H₃BO₃ + NaOH \rightarrow NaBO₂ + 2H₂O \] ...
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Boric acid can be titrated against sodium hydroxide using phenolphthalein as an indicator only in the presence of polyhydroxy compounds like glycol. Glycerol etc. Give reason.

Aqueous solution of orthoboric acid can be titrated against sodium hydroxide using phenolphthalein indicator only in presence of

An alkali is titrated against an acid with methyl orange as indicator , which of the following is a correct combination?

Compound (A) on reaction with iodine in the solvent diglyme gives a hydride (B) and hydrogen gas. The product (B) is instantly hydrolysed by water or aqueous alkali forming compound ( C) and liberating hydrogen gas. The compound ( C) in aqueous solution behaves as a week mono basic acid. But in presence of certain organic polyhydroxy compound behaves as a strong monobasic acid. The hydride (B) in air catches fire spontaneously forming oxide which gives coloured beads with transition metal compounds. Aqueous solution of product (C) can be titrated against sodium hydroxide using phenolphthalein indicator only in presence of:

(a). IF both (A) and (R) are correct and (R) is the correct explanation of (A). (b). If both (A) and (R) are correct but (R) is not the correct explanation of (A). (c). If (A) is correct, but (R) is incorrect. (d). If (A) is incorrect, but (R) is correct. (e) if both (A) and (R) are incorrect. Assertion (A): In the titration of Na_(2)CO_(3) with HCl using methyl orange indicator the volume required at the equivalence point is twice that of the acid required using phenolphthalein indicator. Reason (R): 2 " mol of "HCl are required for complete neutralization of one mole of Na_(2)CO_(3) .

Assertion: In the titration of Na_(2)CO_(3) with HCl using methyl orange indicator, the volume of acid required is twice that of the acid required using phenolphthalein as indicaton. Reason: Two moles of HCl are required for the complete neutralisation of one mole of Na_(2)CO_(3) .

Strong acid versus strong base: The principle of conductometric titrations is based on the fact that during the titration, one of the ions is replaced by the other and invariable these two ions differ in the ionic conductivity with the result that thhe conductivity of the solution varies during the course of the titration. take, for example, the titration between a strong acid, say HCl, and a string base, say NaOH before NaOH is added, the conductance of HCl solution has a high value due to the presence of highly mobile hydrogen ions. As NaOH is added, H^(+) ions are replaced by relatively slower moving Na^(+) ions. consequently the conductance of the solution decreases and this continues right upto the equivalence point where the solution contains only NaCl. Beyond the equivalence point, if more of NaOH is added, then the solution contains a excess of the fast moving OH^(-) ions with the result that its conductance is increased ad it condinues to increase as more and more of NaOH is added. If we plot the conductance value versus the amount of NaOH added, we get a curve of the type shown in Fig. The descending portion AB represents the conductances before the equivalence point (solution contains a mixture of acid HCl and the salt NaCl) and the ascending portion CD represents the conductances after the equivalence point (solution contains the salt NaCl and the excess of NaOH). The point E which represent the minium conductance is due to the solution containing only NaCl with no free acid or alkali and thus represents the equivalence point. this point can, however, be obtained by the extrapolation of the lines AB and DC, and therefore, one is not very particular in locating this point expermentally as it is in the case of ordinary acid-base titrations involving the acid-base indicators. Weak acid versus strong base: Let us take specific example of acetic acid being titrated against NaOH . Before the addition of alkali, the solution shows poor conductance due to feeble ionization of acetic acid. Initially the addition of alkali causes not only the replacement of H^(+) by Na^(+) but also suppresses the dissociation of acetic acid due to the common ion Ac^(-) and thus the conductance of the solution decreases in the beginning. but very soon the conductance start increasing as addition of NaOH neutralizes the undissociated HAc to Na^(+)Ac^(-) thus causing the replacement of non-conducting HAc with Strong-conducting electrolyte Na^(+)Ac^(-) . the increase in conductance continuous right up to the equivalence point. Beyond this point conductance increases more rapidly with the addition of NaOH due to the highly conducting OH^(-) ions, the graph near the equivalence point is curved due to the hydrolysis of the salt NaAc . The actual equivalence point can, as usual, be obtained by the extrapolation method. In all these graphs it has been assumed that the volume change due addition of solution from burrette is negnigible, hence volume change of the solution in beaker the conductance of which is measured is almost constant throughout the measurement. Q. The nature of curve obtained for the titration between weak acid versus strong base as described in the above passage will be:

Statement: In an acid-basic titration involving a strong base and a weak acid, methyl orange can be used as an indicator. Explanation: Methyl orange changes its colour in the pH range 3 to 5.