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Chapter 16: Problem 34

Consider two solutions, solution A and solution B. [H \(\left.^{+}\right]\) in solution A is 25 times greater than that in solution \(B\). What is the difference in the pH values of the two solutions?

Short Answer

Expert verified
The difference in the pH values of the two solutions is approximately 1.40.

Step by step solution

01

Write the equation for the hydrogen ion concentration of both solutions

Given that the concentration of hydrogen ions in solution A is 25 times greater than in solution B, we can write the equation: \[ [H^+ _A] = 25[H^+ _B] \]
02

Find the pH of both solutions using the pH formula

Now, we'll use the pH formula to write the pH values of both solutions. The pH formula is: \[ pH = - \log[H^+] \] So for solution A, the pH is: \[pH_A = -\log[H^+ _A] \] And for solution B, the pH is: \[ pH_B = -\log[H^+ _B] \]
03

Substitute the hydrogen ion concentration equation into the pH equations

We can replace the hydrogen ion concentration in the equations of step 2 to solve for the pH values of both solutions. We know that: \[ [H^+ _A] = 25[H^+ _B] \] So the pH equations become: \[ pH_A = -\log(25[H^+ _B]) \] \[ pH_B = -\log[H^+ _B] \]
04

Calculate the difference in pH values

We want to find the difference between the pH values of the two solutions, which can be written as: \[ \Delta pH = pH_A - pH_B \] Now, we can substitute the pH equations from step 3 into the difference equation: \[ \Delta pH = (-\log(25[H^+ _B])) - (-\log[H^+ _B]) \]
05

Use the logarithm properties to simplify the equation and find the difference

Apply the logarithm property, \(\log{a * b} = \log{a} + \log{b}\), to simplify the equation: \[ \Delta pH = (-\log25 -\log[H^+ _B]) - (-\log[H^+ _B]) \] The -\log[H^+ _B] terms cancel each other out, so we have: \[ \Delta pH = -\log25 \] Now, use the calculator to find -\log(25): \[ \Delta pH \approx 1.40 \] So the difference between the pH values of the two solutions is approximately 1.40.

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Most popular questions from this chapter

(a) Given that \(K_{a}\) for cyanic acid is \(3.5 \times 10^{-4}\) and that for hydrofluoric acid is \(6.8 \times 10^{-4},\) which is the stronger acid? (b) Which is the stronger base, the cyanate ion or the fluoride ion? (c) Calculate \(K_{b}\) values for \(\mathrm{NCO}^{-}\) and \(\mathrm{F}^{-}\).

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, \((\mathbf{c}) 10.0 \mathrm{~mL}\) of $0.0105 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\( diluted to \)500.0 \mathrm{~mL},(\mathbf{d})\( a solution formed by mixing \)20.0 \mathrm{~mL}$ of 0.015 \(M \mathrm{Ba}(\mathrm{OH})_{2}\) with \(40.0 \mathrm{~mL}\) of $8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}$.

Which of the following statements is false? (a) An Arrhenius base increases the concentration of \(\mathrm{OH}^{-}\) in water. (b) A Brønsted-Lowry base is a proton acceptor. (c) Water can act as a Brønsted-Lowry acid. (d) Water can act as a Brønsted-Lowry base. (e) Any compound that contains an -OH group acts as a Brønsted-Lowry base.

The odor of fish is due primarily to amines, especially methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\). Fish is often served with a wedge of lemon, which contains citric acid. The amine and the acid react forming a product with no odor, thereby making the less-than-fresh fish more appetizing. Using data from Appendix \(D\), calculate the equilibrium constant for the reaction of citric acid with methylamine, if only the first proton of the citric acid \(\left(K_{a 1}\right)\) is important in the neutralization reaction.

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for (a) $1.5 \times 10^{-3} \mathrm{MSr}(\mathrm{OH})_{2},\( (b) \)2.250 \mathrm{~g}$ of \(\mathrm{LiOH}\) in \(250.0 \mathrm{~mL}\) of solution, \((\mathbf{c}) 1.00\) \(\mathrm{mL}\) of \(0.175 \mathrm{M} \mathrm{NaOH}\) diluted to $2.00 \mathrm{~L},(\mathbf{d})\( a solution formed by adding \)5.00 \mathrm{~mL}$ of \(0.105 \mathrm{M} \mathrm{KOH}\) to \(15.0 \mathrm{~mL}\) of $9.5 \times 10^{-2} \mathrm{MCa}(\mathrm{OH})_{2}$
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