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Chapter 15: Problem 29

Basic solution \(A\) has \(\mathrm{pH}=9 .\) Basic solution \(\mathrm{B}\) is ten times more basic than \(\Lambda\). What is the \(\mathrm{pH}\) of solution \(\mathrm{B}\) ?

Short Answer

Expert verified
The pH of solution B is 10.

Step by step solution

01

Find the pOH of solution A

First, we need to find the pOH of solution A. We can use the relationship between pH and pOH, which is given by the formula: pOH = 14 - pH For solution A, we have pH = 9. Plugging this value into the formula, we get: pOH = 14 - 9 = 5 So, the pOH of solution A is 5.
02

Determine the pOH of solution B

We are given that solution B is ten times more basic than solution A. Since the pOH scale is logarithmic, this factor of 10 corresponds to a change of 1 unit in the pOH value. To find the pOH of solution B, we subtract 1 from the pOH of solution A: pOH(B) = pOH(A) - 1 With pOH(A) = 5, we get: pOH(B) = 5 - 1 = 4 So, the pOH of solution B is 4.
03

Calculate the pH of solution B

Finally, we can find the pH of solution B using the relationship between pH and pOH: pH = 14 - pOH With pOH(B) = 4, we obtain: pH(B) = 14 - 4 = 10 Therefore, the pH of solution B is 10.

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Most popular questions from this chapter

A solution is prepared by dissolving \(2.00\) moles of \(\mathrm{HNO}_{3}\) in enough water to get \(800.0 \mathrm{~mL}\) of solution. What are the \(\mathrm{H}_{3} \mathrm{O}^{+}\) and the \(\mathrm{OH}\) molar concentrations?

For each acid-base pair, write a balanced equation for the neutralization reaction: (a) Lithium hydroxide and hydroiodic acid (b) Acetic acid and sodium hydroxide (c) Hydrobromic acid and calcium hydroxide (d) Potassium hydroxide and phosphoric acid (assume all acidic protons react)

If you dissolve \(0.250 \mathrm{~g}\) of \(\mathrm{Ba}(\mathrm{OH})_{2}\) in \(3.00 \mathrm{~L}\) of water, what is the \(\mathrm{pH}\) of the solution?

Without using a calculator, what is the base- 10 logarithm of \(10^{0}\) ? Of 1?

True or false? In an aqueous solution at \(25^{\circ} \mathrm{C}\), you will always get the same number when you multiply the equilibrium \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration by the equilibrium OH \(^{-}\) concentration. Explain your answer.
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