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Chapter 16: Problem 64

Suppose you wanted to produce an aqueous solution of \(\mathrm{pH}=8.65\) by dissolving one of the following salts in water. Which salt would you use, and at what molarity? (a) \(\mathrm{NH}_{4} \mathrm{Cl} ;\) (b) \(\mathrm{KHSO}_{4} ;\) (c) \(\mathrm{KNO}_{2}\); (d) \(\mathrm{NaNO}_{3}\).

Short Answer

Expert verified
The correct salt to use to produce an aqueous solution with a pH of 8.65 would be KNO2. Its molarity should be calculated as per the steps given.

Step by step solution

01

Identify the Dissociation in Water

The initial step is to write out the dissociation in water for each of the given salts. For example, NH4Cl dissociates into NH4+ and Cl-. This is done by breaking the salts into their constituent ions.
02

Determine the Resulting Solution from the Dissociation

Now, figure out whether the solution resulting from the dissociation of each salt is acidic, neutral, or basic. The NH4+ ion can donate a proton to become NH3, making the solution acidic. Cl- is the conjugate base of a strong acid (HCl), which does not accept protons readily, hence does not contribute to pH. This process is performed for each salt, resulting in the identification of the potential pH of a solution they form.
03

Select the Correct Salt

After determining the properties of the salts, one can select from them the one that gives a solution with a pH of 8.65, which is a slightly basic pH. From the previous steps, it is evident that only KNO2 and NaN03 can create a basic solution. However, since NaNO3 is a salt of a strong base (NaOH) and a strong acid (HNO3), it will not affect the pH of the solution. Thus, the correct salt that can provide the desired pH is KNO2.
04

Calculate Molarity of the Selected Salt

Following the selection of KNO2 as the correct salt, the next step is to calculate its molarity to achieve the desired pH. This necessitates the use of the equation \(pOH = -\log[OH^-]\) to find the pOH first (since pH + pOH = 14 at 25°C), and then apply \(pOH = pKa + \log\frac{[A-]}{[HA]}\) (since only \(K_a\) for HNO2 is readily available), where [A-]=molarity of KNO2 and [HA] is an insignificant 0. Calculate for [A-] to get the molarity of KNO2.

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Most popular questions from this chapter

Pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\left(\mathrm{p} K_{\mathrm{b}}=8.82\right),\) forms a salt, pyridinium chloride, as a result of a reaction with HCl. Write an ionic equation to represent the hydrolysis of the pyridinium ion, and calculate the \(\mathrm{pH}\) of \(0.0482 \mathrm{M} \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} \mathrm{Cl}^{-}(\mathrm{aq})\).

Complete the following equations in those instances in which a reaction (hydrolysis) will occur. If no reaction occurs, so state. (a) \(\mathrm{NH}_{4}^{+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \longrightarrow\) (b) \(\mathrm{Na}^{+}(\mathrm{aq})+\mathrm{NO}_{2}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \longrightarrow\) (c) \(\mathrm{K}^{+}(\mathrm{aq})+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \longrightarrow\) (d) \(\mathrm{K}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Na}^{+}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq})+\) (e) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2} \mathrm{O} \longrightarrow\)

What is the pH of each of the following solutions? (a) \(0.0045 \mathrm{M} \mathrm{HCl} ;\) (b) \(6.14 \times 10^{-4} \mathrm{M} \mathrm{HNO}_{3} ;\) (c) 0.00683 \(\mathrm{M} \mathrm{NaOH} ;\) (d) \(4.8 \times 10^{-3} \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\)

Explain why \(\left[\mathrm{PO}_{4}^{3-}\right]\) in \(1.00 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) is not simply \(\frac{1}{3}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\) but much, much less than \(\frac{1}{3}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\)

The reaction of \(\mathrm{CH}_{3} \mathrm{COOH}(\) aq) proceeds furthest toward completion with a base when that base is (a) \(\mathrm{H}_{2} \mathrm{O} ;\) (b) \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+} ;\) (c) \(\mathrm{NH}_{4}^{+} ;\) (d) \(\mathrm{Cl}^{-} ;\) (e) \(\mathrm{CO}_{3}^{2-}\).
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