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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

Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, and HCl. Milk of Magnesia, a suspension of solid Mg(OH)2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs

What two common assumptions can simplify the calculation of equilibrium concentrations in a solution of a weak base?

What will be the\(pH\)of a buffer solution prepared from\(0.20\;mol N{H_3}, 0.40\;mol N{H_4}N{O_3}\), and just enough water to give\(1.00\;L\)of solution?

The \(pH\) of a solution of household ammonia, a \(0.950M\) solution of \(N{H_3}\), is \(11.612.\) Determine \({K_b} \)for \(N{H_3}\) from these data.

From the equilibrium concentrations given, calculate for each of the weak acids and for each of the weak bases.

\(\begin{aligned}(a)C{H_3}C{O_2}H:\left( {{H_3}{O^ + }} \right) = 1.34 \times 1{0^{ - 3}}M;\left( {C{H_3}CO_2^ - } \right) = 1.34 \times 1{0^{ - 3}}M;\left( {C{H_3}C{O_2}H} \right) = 9.866 \times 1{0^{ - 2}}M;\\(b)Cl{O^ - }:\left( {O{H^ - }} \right) = 4.0 \times 1{0^{ - 4}}M;(HClO) = 2.38 \times 1{0^{ - 5}}M;\left( {Cl{O^ - }} \right) = 0.273M;\\(c)HC{O_2}H:\left( {HC{O_2}H} \right) = 0.524M;\left( {{H_3}{O^ + }} \right) = 9.8 \times 1{0^{ - 3}}M\left( {HCO_2^ - } \right) = 9.8 \times 1{0^{ - 3}}M;\\(d){C_6}{H_5}NH_3^ + :\left( {{C_6}{H_5}NH_3^ + } \right) = 0.233M;\left( {{C_6}{H_5}N{H_2}} \right) = 2.3 \times 1{0^{ - 3}}M;\left( {{H_3}{O^ + }} \right) = 2.3 \times 1{0^{ - 3}}M\end{aligned}\)
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