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Chapter 16: Problem 71

A certain salt, MX (containing the \(\mathrm{M}^{+}\) and \(\mathrm{X}^{-}\) ions), is dissolved in water, and the \(\mathrm{pH}\) of the resulting solution is 7.0. Can you say anything about the strengths of the acid and the base from which the salt is derived?

Short Answer

Expert verified
The given salt MX is derived from a strong acid and a strong base since its solution in water is neutral (pH = 7). This is because a strong acid and strong base neutralize each other's effects, leading to a neutral pH. Thus, neither the \(\mathrm{M}^{+}\) cation nor \(\mathrm{X}^{-}\) anion exhibits acidic or basic behaviour.

Step by step solution

01

Understand pH and its relation to acids and bases

The pH scale is used to determine how acidic or alkaline a water-based solution is. Acidic solutions have lower pH values (0-6), basic solutions have higher pH values (8-14), and 7 is neutral. The given pH value is 7.0, implying the solution is neutral. Therefore, the solution doesn't favor acidity or basicity.
02

Becoming familiar with the relationship between salts and their parent acid and base

Acid-base neutralization forms a salt. If a salt is formed by neutralizing a strong acid with a strong base, it doesn't affect the pH of the solution when dissolved in water as it’s neutral. If a salt is produced by a strong acid and weak base, the solution will be acidic when dissolved. On the other hand, if it's from a weak acid and strong base, the resulting solution will be basic.
03

Identify the strengths of the acid and base

As the solution of the salt MX is neutral (pH 7.0), it implies the salt is derived from the neutralization of a strong acid by a strong base. The reasoning is that a strong acid and a strong base neutralize each other's effects leading to a neutral solution. Neither the \(\mathrm{M}^{+}\) cation nor \(\mathrm{X}^{-}\) anion acts as an acid or base in water, that's why the solution is not acidic or basic.

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Most popular questions from this chapter

Classify these following oxides as acidic, basic, amphoteric, or neutral: (a) \(\mathrm{CO}_{2},\) (b) \(\mathrm{K}_{2} \mathrm{O},\) (c) \(\mathrm{CaO}\) (d) \(\mathrm{N}_{2} \mathrm{O}_{5},(\mathrm{e}) \mathrm{CO},(\mathrm{f}) \mathrm{NO}\) \((\mathrm{g}) \mathrm{SnO}_{2},(\mathrm{~h}) \mathrm{SO}_{3},(\mathrm{i}) \mathrm{Al}_{2} \mathrm{O}_{3}\) (j) \(\mathrm{BaO}\).

Identify the acid-base conjugate pairs in each of these reactions: (a) \(\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{HCN} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{CN}^{-}\) (b) \(\mathrm{HCO}_{3}^{-}+\mathrm{HCO}_{3}^{-} \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}+\mathrm{CO}_{3}^{2-}\) (c) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{NH}_{3} \rightleftharpoons \mathrm{HPO}_{4}^{2-}+\mathrm{NH}_{4}^{+}\) (d) \(\mathrm{HClO}+\mathrm{CH}_{3} \mathrm{NH}_{2} \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}+\mathrm{ClO}^{-}\) (e) \(\mathrm{CO}_{3}^{2-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{HCO}_{3}^{-}+\mathrm{OH}^{-}\) (f) \(\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{OH}^{-}\)

The \(\mathrm{pH}\) of a solution is 6.7 . From this statement alone, can you conclude that the solution is acidic? If not, what additional information would you need? Can the \(\mathrm{pH}\) of a solution be zero or negative? If so, give examples to illustrate these values.

Predict whether a solution containing the salt \(\mathrm{K}_{2} \mathrm{HPO}_{4}\) will be acidic, neutral, or basic. (Hint: You need to consider both the ionization and hydrolysis of \(\mathrm{HPO}_{4}^{2-} .\)

A \(0.400 M\) formic acid (HCOOH) solution freezes at \(-0.758^{\circ} \mathrm{C} .\) Calculate the \(K_{\mathrm{a}}\) of the acid at that temperature. (Hint: Assume that molarity is equal to molality. Carry your calculations to three significant figures and round off to two for \(K_{\mathrm{a}}\) .)
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