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Chapter 15: Problem 98

Determine the concentration of a \(\mathrm{NH}_{4} \mathrm{Cl}\) solution that has a pH of 5.64 .

Short Answer

Expert verified
The concentration of \( NH_4Cl \) solution is equal to the \( H_3O^+ \) concentration, which is \(10^{-5.64}\) M.

Step by step solution

01

Convert pH to hydronium ion concentration

Using the formula \(pH = -log[H_3O^+]\), we can determine the concentration of \(H_3O^+\) from the given pH. Here, \(H_3O^+\) concentration = \(10^{-pH}\). Given pH = 5.64, \(H_3O^+\) concentration = \(10^{-5.64}\). Evaluate the expression to find \(H_3O^+\) concentration in \(M\).
02

Write the dissociation equation

The equation for the dissociation of \(NH_4^+\) into \(NH_3\) and \(H^+\) in water is: \(NH_4^+ \rightleftharpoons NH_3 + H^+\). As \(NH_4^+\) donates one proton, the concentration of \(NH_4Cl\) equals to the concentration of \(H_3O^+\).
03

Find the concentration of \(NH_4Cl\)

Since the concentration of \(NH_4Cl\) equals the concentration of \(H_3O^+\), the molarity of \(NH_4Cl\) is the same as the \(H_3O^+\) concentration we calculated in step 1.

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Most popular questions from this chapter

What are the Lewis definitions of an acid and a base? In what way are they more general than the Brønsted definitions?

What is the strongest acid and the strongest base that can exist in water?

Which of the following is a stronger base: \(\mathrm{NH}_{3}\) or \(\mathrm{PH}_{3} ?\) (Hint: The \(\mathrm{N}-\mathrm{H}\) bond is stronger than the \(\mathrm{P}-\mathrm{H}\) bond. \()\)

Explain why small, highly charged metal ions are able to undergo hydrolysis.

A \(0.400 M\) formic acid (HCOOH) solution freezes at \(-0.758^{\circ} \mathrm{C}\). Calculate the \(K_{\mathrm{a}}\) of the acid at that temperature. (Hint: Assume that molarity is equal to molality. Carry your calculations to three significant figures and round off to two for \(K_{\mathrm{a}}\).)
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