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Chapter 14: Q2E (page 826)

Question: Write equations that show \({H_2}PO_4^ - \) acting both as an acid and as a base.

Short Answer

Expert verified

When \({H_2}PO_4^ - \) reacts with \({H_2}O\), it gives proton to water. As a result, it acts as an acid. The equation regarding this is as follows

\({H_2}PO_4^ - + {H_2}O \to HPO_4^{2 - } + {H_3}{O^ + }\)

When \({H_2}PO_4^ - \) reacts with \(HCl\), it takes proton from \(HCl\). As a result, it acts as a base.

The equation regarding this is as follows

\({H_2}PO_4^ - + HCl \to {H_3}P{O_4} + C{l^ - }\)

Step by step solution

01

Define the Bronsted-Lowry concept of acid and base.

The Bronsted-lowry concept of acid and base tells that an acid is a proton\(({H^ + })\) donor, and a base is a proton acceptor.

When a Bronsted-Lowry acid loses a proton, a conjugate base is formed. Similarly, when a Bronsted-Lowry base gains a proton, a conjugate acid is formed.

02

\({H_2}PO_4^ - \) as an acid.

\({H_2}PO_4^ - + {H_2}O \to HPO_4^{2 - } + {H_3}{O^ + }\)

Here, \({H_2}PO_4^ - \) is giving proton to water, so it is an acid.

03

\({H_2}PO_4^ - \) as a base.

\({H_2}PO_4^ - + HCl \to {H_3}P{O_4} + C{l^ - }\)

Here, \({H_2}PO_4^ - \) is taking proton, so it is a base.

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Most popular questions from this chapter

Draw a curve for a series of solutions of HF. Plot \({\left[ {{H_3}{O^ + }} \right]_{total }}\) on the vertical axis and the total concentration of HF (the sum of the concentrations of both the ionized and nonionized HF molecules) on the horizontal axis. Let the total concentration of HF vary from \(1 \times 1{0^{ - 10}}M\) to\(1 \times 1{0^{ - 2}}M\) .

Which of the following will increase the percentage of HF that is converted to the fluoride ion in water?

(a) Addition of \(NaOH\)

(b) Addition of \(HCl\)

(c) Addition of \(NaF\)

Novocaine, \({C_{13}}{H_{21}}{O_2}\;{N_2}Cl\), is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is \(7 \times 1{0^{ - 6}}\). Is a solution of novocaine acidic or basic? What are \(\left( {{H_3}{O^ + }} \right),\left( {O{H^ - }} \right)\), and \(pH\) of a \(2.0\% \) solution by mass of novocaine, assuming that the density of the solution is \(1.0\;g/mL\).

The active ingredient formed by aspirin in the body is salicylic acid, \({C_6}{H_4}OH\left( {C{O_2}H} \right)\). The carboxyl group \(\left( { - C{O_2}H} \right)\)acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a \(0.001M\) aqueous solution of \({C_6}{H_4}OH\left( {C{O_2}H} \right)\)

A \({\bf{5}}{\bf{.36 g}}\) sample of \({\bf{N}}{{\bf{H}}_{\bf{4}}}{\bf{Cl}}\) was added to \({\bf{25}}.{\bf{0}}{\rm{ }}{\bf{mL}}\) of \({\bf{1}}{\bf{.00 M NaOH}}\) and the resulting solution diluted to\({\bf{0}}.{\bf{100}}{\rm{ }}{\bf{L}}\).

(a) What is the pH of this buffer solution?

(b) Is the solution acidic or basic?

(c) What is the pH of a solution that results when \({\bf{3}}.{\bf{00}}{\rm{ }}{\bf{mL}}\) of \({\bf{0}}.{\bf{034}}{\rm{ }}{\bf{M}}{\rm{ }}{\bf{HCl}}\) is added to the solution?
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