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Chapter 15: Problem 27

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

Short Answer

Expert verified
The pH of the solution of 1.00 M HNO₂ and 1.00 M NaNO₂ is approximately 3.35.

Step by step solution

01

Identify the initial concentrations

We are given a solution that is 1.00 M in both HNO₂ and NaNO₂. To calculate the pH, we need to consider the dissociation of these molecules in water. For the weak acid HNO₂, the dissociation reaction is: \[ HNO_{2}(aq) \rightleftharpoons H^{+}(aq) + NO_{2}^{-}(aq) \] For NaNO₂, the dissociation reaction is: \[ NaNO_{2}(aq) \rightarrow Na^{+}(aq) + NO_{2}^{-}(aq) \]
02

Determine the equilibrium concentrations

Since NaNO₂ is a strong electrolyte, it will dissociate completely in the solution, providing 1.00 M of NO₂⁻ ions initially. HNO₂, being a weak acid, will only partially dissociate. Let's call the extent of dissociation x. At equilibrium, the concentration of HNO₂ will be (1.00 - x) M, while the concentrations of H⁺ and NO₂⁻ will be x and (1+x) M, respectively.
03

Apply the Ka value of HNO₂

The dissociation constant Ka (the acid ionization constant) for HNO₂ can be found in a table of Ka values. The value for HNO₂ is: \[ K_{a} = 4.50 \times 10^{-4} \] Using the expression for the Ka, we can establish a relationship between the concentrations of the species at equilibrium: \[ K_{a} = \frac{[H^{+}][NO_{2}^{-}]}{[HNO_{2}]} \] Substituting the equilibrium concentrations, we get: \[ 4.50 \times 10^{-4} = \frac{(x)(1+x)}{(1-x)} \]
04

Solve for x

Since Ka is a small value, we can approximate by assuming x is very small compared to 1 and thus: \[ 4.50 \times 10^{-4} \approx \frac{x}{(1-x)} \approx \frac{x}{1} \] Solving for x, we find: \[ x \approx 4.50 \times 10^{-4} \]
05

Use the Henderson-Hasselbalch equation

Now that we have the concentrations of H⁺, HNO₂, and NO₂⁻ at equilibrium, we can use the Henderson-Hasselbalch equation to find the pH of the solution: \[ pH = pK_{a} + \log \frac{[A^{-}]}{[HA]} \] Where pKₐ = -log(Kₐ), and [A⁻] and [HA] are the equilibrium concentrations of the conjugate base and the weak acid, respectively. Plugging in the values: \[ pH = -\log(4.50 \times 10^{-4}) + \log \frac{(1+4.50 \times 10^{-4})}{(1-4.50 \times 10^{-4})} \]
06

Calculate the pH

Solve the equation to find the pH: \[ pH \approx 3.35 \] The pH of the solution of 1.00 M HNO₂ and 1.00 M NaNO₂ is approximately 3.35.

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Most popular questions from this chapter

Calculate the pH of each of the following buffered solutions. a. \(0.10 M\) acetic acid \(/ 0.25 M\) sodium acetate b. \(0.25 M\) acetic acid \(/ 0.10 M\) sodium acetate c. \(0.080 M\) acetic acid \(/ 0.20 M\) sodium acetate d. \(0.20 M\) acetic acid \(0.080 M\) sodium acetate

The active ingredient in aspirin is acetylsalicylic acid. A \(2.51-g\) sample of acetylsalicylic acid required \(27.36 \mathrm{~mL}\) of \(0.5106 M\) \(\mathrm{NaOH}\) for complete reaction. Addition of \(13.68 \mathrm{~mL}\) of \(0.5106 \mathrm{M}\) \(\mathrm{HCl}\) to the flask containing the aspirin and the sodium hydroxide produced a mixture with \(\mathrm{pH}=3.48\). Determine the molar mass of acetylsalicylic acid and its \(K_{\mathrm{a}}\) value. State any assumptions you must make to reach your answer.

Consider the following four titrations. i. \(100.0 \mathrm{~mL}\) of \(0.10 M \mathrm{HCl}\) titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) ii. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{NaOH}\) titrated by \(0.10 \mathrm{M} \mathrm{HCl}\) iii. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\) titrated by \(0.10 \mathrm{M} \mathrm{HCl}\) iv. \(100.0 \mathrm{~mL}\) of \(0.10 M\) HF titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) Rank the titrations in order of: a. increasing volume of titrant added to reach the equivalence point. b. increasing pH initially before any titrant has been added. c. increasing \(\mathrm{pH}\) at the halfway point in equivalence. d. increasing \(\mathrm{pH}\) at the equivalence point. How would the rankings change if \(\mathrm{C}_{5} \mathrm{H}_{3} \mathrm{~N}\) replaced \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) and if \(\mathrm{HOC}_{6} \mathrm{H}_{5}\) replaced \(\mathrm{HF}\) ?

Consider a solution containing \(0.10 M\) ethylamine \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\right.\) ), \(0.20 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{NH}_{3}^{+}\), and \(0.20 \mathrm{M} \mathrm{Cl}^{-}\) a. Calculate the \(\mathrm{pH}\) of this solution. b. Calculate the \(\mathrm{pH}\) after \(0.050 \mathrm{~mol} \mathrm{KOH}(s)\) is added to \(1.00 \mathrm{~L}\) of this solution. (Ignore any volume changes.)

A certain acetic acid solution has \(\mathrm{pH}=2.68\). Calculate the volume of \(0.0975 M \mathrm{KOH}\) required to reach the equivalence point in the titration of \(25.0 \mathrm{~mL}\) of the acetic acid solution.
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