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Chapter 15: Problem 35

Calculate the \(\mathrm{pH}\) of a solution that is 1.00 \(\mathrm{M}\) HNO, and 1.00 \(\mathrm{M} \mathrm{NaNO}_{2}\)

Short Answer

Expert verified
Using the Henderson-Hasselbalch equation, the pH of a solution containing 1.00 M HNO and 1.00 M NaNO₂ can be calculated as follows: \( pH = pKa + \log \frac{[Conjugate \, Base]}{[Acid]} \). Assuming a pKa value of 3.5 for HNO, the calculation becomes \( pH = 3.5 + \log \frac{1.00}{1.00} \), which simplifies to \( pH = 3.5 \). Therefore, the pH of the solution is 3.5.

Step by step solution

01

Identify the given values

In this problem, the given concentrations of the acid (HNO) and its conjugate base (NO₂⁻) are both 1.00 M. We will use these values in the Henderson-Hasselbalch equation to calculate the pH of the solution.
02

Write down the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation relates the pH, pKa, and the concentrations of the weak acid and its conjugate base as follows: \[ pH = pKa + \log \frac{[Conjugate \, Base]}{[Acid]} \]
03

Determine the pKa of HNO

To use the Henderson-Hasselbalch equation, we need the pKa value of HNO. Unfortunately, the pKa value of HNO is not commonly known or found in most tables. In such cases, either the pKa value must be given in the problem or it should be approximated. For the sake of this exercise, let's assume the given pKa value for HNO is 3.5, which is a reasonable approximation for a weak acid.
04

Use the given concentrations and pKa in the Henderson-Hasselbalch equation

Now, we can use the given concentrations of the acid (HNO) and its conjugate base (NO₂⁻), as well as the assumed pKa value of HNO, in the Henderson-Hasselbalch equation: \[ pH = 3.5 + \log \frac{1.00}{1.00} \]
05

Calculate the pH of the solution

Since the concentrations of the conjugate base and the acid are equal, the log term in the equation becomes 0. Therefore, the pH of the solution is the same as the pKa value: \[ pH = 3.5 + 0 \] Thus, the pH of the solution containing 1.00 M HNO and 1.00 M NaNO₂ is 3.5.

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Most popular questions from this chapter

Calculate the number of moles of \(\mathrm{HCl}(g)\) that must be added to 1.0 \(\mathrm{L}\) of 1.0 $\mathrm{M} \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ to produce a solution buffered at each pH. $$ \text{(a)}\mathrm{pH}=\mathrm{p} K_{\mathrm{a}} \quad \text { b. } \mathrm{pH}=4.20 \quad \text { c. } \mathrm{pH}=5.00 $$

You make 1.00 \(\mathrm{L}\) of a buffered solution \((\mathrm{pH}=4.00)\) by mixing acetic acid and sodium acetate. You have 1.00\(M\) solutions of each component of the buffered solution. What volume of each solution do you mix to make such a buffered solution?

Sketch the titration curve for the titration of a generic weak base \(\mathrm{B}\) with a strong acid. The titration reaction is $$ \mathrm{B}+\mathrm{H}^{+} \rightleftharpoons \mathrm{BH}^{+} $$ On this curve, indicate the points that correspond to the following: a. the stoichiometric (equivalence) point b. the region with maximum buffering c. \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}\) d. \(\mathrm{pH}\) depends only on \([\mathrm{B}]\) e. \(\mathrm{pH}\) depends only on \(\left[\mathrm{BH}^{+}\right]\) f. \(\mathrm{pH}\) depends only on the amount of excess strong acid added

A 0.210 -g sample of an acid (molar mass \(=192 \mathrm{g} / \mathrm{mol}\) ) is titrated with 30.5 \(\mathrm{mL}\) of 0.108\(M \mathrm{NaOH}\) to a phenolphthalein end point. Is the acid monoprotic, diprotic, or triprotic?

Calculate the pH after 0.010 mole of gaseous HCl is added to 250.0 \(\mathrm{mL}\) of each of the following buffered solutions. $$ \begin{array}{l}{\text { a. } 0.050 M \mathrm{NH}_{3} / 0.15 \mathrm{MNH}_{4} \mathrm{Cl}} \\ {\text { b. } 0.50 \mathrm{M} \mathrm{NH}_{3} / 1.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}}\end{array} $$ Do the two original buffered solutions differ in their pH or their capacity? What advantage is there in having a buffer with a greater capacity?
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