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Chapter 19: Q19.47P (page 874)

How does the titration curve of a monoprotic acid differ from that of a diprotic acid?

Short Answer

Expert verified

The two breaks in a diprotic acid titration curve correspond to the two locations where the pH rises quickly. In a monoprotic acid, there is only one break.

Step by step solution

01

Definition of equilibrium

A chemical equilibrium is a state in which no net change in the amount of reactants and products occur during a reversible chemical reaction.

02

Step 2: How the titration curve of a monoprotic acid differ

A diprotic acid titration curve has two breaks, which reflect the two regions, where the rises rapidly. But in a monoprotic acid, there is only one break.

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Most popular questions from this chapter

Scenes A to C represent aqueous solutions of the slightly soluble salt MZ (only the ions of this salt are shown):

$MZ (s) ⇌ M^{2 +} (aq) + Z^{2 -} (aq)$

(a) Which scene represents the solution just after solid MZ is stirred thoroughly in distilled water?

(b) If each sphere represents $2.5 \times 10^{- 6} M$ of ions, what is the $K_{sp}$ of MZ?

(c) Which scene represents the solution after ${Na}_{2} Z (aq)$ is added?

(d) If $Z^{2 -}$ is $C {O_{3}}^{2 -}$ , which scene represents the solution after thehas been lowered?

Which compound in each pair is more soluble in water?

(a) Manganese(II)hydroxide or calcium iodate

(b) Strontium carbonate or cadmium sulfide

(c) Silver cyanide or copper(I)iodide

Explain how strong acid–strong base, weak acid–strong base, and weak base–strong acid titrations using the same concentrations differ in terms of (a) the initial pH and (b) the pH at the equivalence point. (The component in italics is in the flask.)

Ethylenediaminetetraacetic acid (abbreviated $H_{4} EDTA$ ) is a tetraprotic acid. Its salts are used to treat toxic metal poisoning by forming soluble complex ions that are then excreted. Because ${EDTA}^{4 -}$ also binds essential calcium ions, it is often administered as the calcium disodium salt. For example, when ${Na}_{2} Ca$ (EDTA) is given to a patient ${[Ca (EDTA)]}^{2 -}$ , the ions react with circulating ions and the metal ions are exchanged:

$\begin{array}{l} {[Ca (EDTA)]}^{2 -} (aq) + P b^{2 +} (aq) \\ {[Pb (EDTA)]}^{2 -} (aq) + C a^{2 +} (aq) K_{c} = 2.5 \times 10^{7} \end{array}$

A child has a dangerous blood lead level of $120 μg / 100 mL$ . If the child is administered $100 mL$ of $0.10 M {NaNa}_{2}$ (EDTA), assuming the exchange reaction and excretion process are $100 %$ efficient, what is the final concentration of ${Pb}^{2 +}$ in $μg / 100 mL$ blood? (Total blood volume is $1.5 L$ .)

The indicator cresol red has $K_{a} = 3.5 \times 10^{- 9}$ . Over what approximate pH range does it change color?

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