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Chapter 16: Problem 17

Designate the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each of the following equations, and also designate the conjugate acid and conjugate base of each on the right side: $$ \text { (a) } \mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q) $$ (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ (c) \(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons\) $$ \mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q) $$

Short Answer

Expert verified
(a) The Bronsted-Lowry acid is \(\mathrm{NH}_{4}^{+}\), the Bronsted-Lowry base is \(\mathrm{CN}^{-}\), the conjugate acid is \(\mathrm{HCN}\), and the conjugate base is \(\mathrm{NH}_{3}\). (b) The Bronsted-Lowry acid is \((\mathrm{CH}_{3})_{3} \mathrm{~N}\), the Bronsted-Lowry base is \(\mathrm{H}_{2}\mathrm{O}\), the conjugate acid \((\mathrm{CH}_{3})_{3} \mathrm{NH}^{+}\), and the conjugate base is \(\mathrm{OH}^{-}\). (c) The Bronsted-Lowry acid is \(\mathrm{HCOOH}\), the Bronsted-Lowry base is \(\mathrm{PO}_{4}^{3-}\), the conjugate acid is \(\mathrm{HPO}_{4}^{2-}\), and the conjugate base is \(\mathrm{HCOO}^{-}\).

Step by step solution

01

(a) Identify Bronsted-Lowry acid and base on the left side

In the given equation, \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q),\) we can see that \(\mathrm{NH}_{4}^{+}\) donates a proton to \(\mathrm{CN}^{-}.\) So, the Bronsted-Lowry acid is \(\mathrm{NH}_{4}^{+}\), and the Bronsted-Lowry base is \(\mathrm{CN}^{-}.\)
02

(a) Identify the conjugate acid and base on the right side

On the right side of the equation, after \(\mathrm{NH}_{4}^{+}\) donates a proton, it forms \(\mathrm{NH}_{3}.\) Therefore, the conjugate base is \(\mathrm{NH}_{3}.\) After receiving a proton, \(\mathrm{CN}^{-}\) becomes \(\mathrm{HCN}.\) Therefore, the conjugate acid is \(\mathrm{HCN}.\)
03

(b) Identify Bronsted-Lowry acid and base on the left side

In this equation, \((\mathrm{CH}_{3})_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons (\mathrm{CH}_{3})_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q),\) we can see that \((\mathrm{CH}_{3})_{3} \mathrm{~N}\) donates a proton to \(\mathrm{H}_{2}\mathrm{O}.\) So, the Bronsted-Lowry acid is \((\mathrm{CH}_{3})_{3} \mathrm{~N},\) and the Bronsted-Lowry base is \(\mathrm{H}_{2}\mathrm{O}.\)
04

(b) Identify the conjugate acid and base on the right side

On the right side of the equation, after donating a proton, \((\mathrm{CH}_{3})_{3} \mathrm{~N}\) forms \((\mathrm{CH}_{3})_{3} \mathrm{NH}^{+}.\) Therefore, the conjugate acid is \((\mathrm{CH}_{3})_{3} \mathrm{NH}^{+}.\) After receiving a proton, \(\mathrm{H}_{2}\mathrm{O}\) becomes \(\mathrm{OH}^{-}.\) Therefore, the conjugate base is \(\mathrm{OH}^{-}.\)
05

(c) Identify Bronsted-Lowry acid and base on the left side

In this equation, \(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons \mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q),\) we can see that \(\mathrm{HCOOH}\) donates a proton to \(\mathrm{PO}_{4}^{3-}.\) So, the Bronsted-Lowry acid is \(\mathrm{HCOOH},\) and the Bronsted-Lowry base is \(\mathrm{PO}_{4}^{3-}.\)
06

(c) Identify the conjugate acid and base on the right side

On the right side of the equation, after donating a proton, \(\mathrm{HCOOH}\) forms \(\mathrm{HCOO}^{-}.\) Therefore, the conjugate base is \(\mathrm{HCOO}^{-}.\) After receiving a proton, \(\mathrm{PO}_{4}^{3-}\) becomes \(\mathrm{HPO}_{4}^{2-}.\) Therefore, the conjugate acid is \(\mathrm{HPO}_{4}^{2-}.\)

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Most popular questions from this chapter

Explain the following observations: (a) HCl is a stronger acid than \(\mathrm{H}_{2} \mathrm{~S} ;\) (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is a stronger acid than \(\mathrm{H}_{3} \mathrm{AsO}_{4} ;\) (c) \(\mathrm{HBrO}_{3}\) is a stronger acid than \(\mathrm{HBrO}_{2}\); (d) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) is a stronger acid than \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-} ;(\mathrm{e})\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) is a stronger acid than phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\).

(a) Why is \(\mathrm{NH}_{3}\) a stronger base than \(\mathrm{H}_{2} \mathrm{O} ?\) (b) Why is \(\mathrm{NH}_{3}\) a stronger base than \(\mathrm{CH}_{4} ?\)

(a) Which of the following is the stronger Bronsted-Lowry acid, HBrO or HBr? (b) Which is the stronger Bronsted-Lowry base, \(\mathrm{F}^{-}\) or \(\mathrm{Cl}^{-}\) ? Briefly explain your choices.

(a) Give the conjugate base of the following Bronsted-Lowry acids: (i) \(\mathrm{HCOOH},\) (ii) \(\mathrm{HPO}_{4}^{2-}\). (b) Give the conjugate acid of the following Bronsted-Lowry bases: (i) \(\mathrm{SO}_{4}^{2-}\), (ii) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\)

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: (a) \(0.182 \mathrm{M} \mathrm{KOH},\) (b) \(3.165 \mathrm{~g}\) of \(\mathrm{KOH}\) in \(500.0 \mathrm{~mL}\) of solution, (c) \(10.0 \mathrm{~mL}\) of \(0.0105 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) diluted to \(500.0 \mathrm{~mL}\) (d) a solution formed by mixing \(20.0 \mathrm{~mL}\) of \(0.015 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) with \(40.0 \mathrm{~mL}\) of \(8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}\).
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