8.1 Introduction to Acids and Bases

Multiple Choice

A Bronsted Lowrey acid is a...

• proton donor

• proton accepter

Match

Match the following

• • CH₃CO₂H(aq)+H₂O(l) ⇋ CH₃CO₂⁻(aq) + H₃O⁺(aq)

• CH₃CO₂H

• H₂O

• CH₃CO₂⁻

• H₃O⁺

acid

base

conjugate base

conjugate acid

• acid

• base

• conjugate base

• conjugate acid

Match

Match the following

• • CH₃CO₂H(aq)+H₂O(l) ⇋ CH₃CO₂⁻(aq) + H₃O⁺(aq)

• CH₃CO₂H

• H₂O

• CH₃CO₂⁻

• H₃O⁺

acid

base

conjugate base

conjugate acid

• acid

• base

• conjugate base

• conjugate acid

Multiple Choice

If the pH of a solution is 4.0, what is the pOH?

• 4.0

• 10.0

• 1.0 x 10 ^-4

• cannot be determined from the information

Multiple Choice

The pH of a solution is 8.43. What is the [H₃O⁺]concentration?

• 3.7 x 10 ^-9

• 1.0 x 10^-8.43

• 2.7 x 10^-6

• 1.0 x 10^-14

Possible Quiz Questions

Multiple Choice

In pure water, some of the molecules ionize according to the equation H₂O <--> H⁺ + OH^-. The extent of the ionization increases with temperature. The graph above shows the pH of pure water at different temperature. Which of the following represents the variations in the pOH of pure water under the same conditions?

Multiple Choice

H₂O (l) + H₂O (l) <--> H₃O⁺ + OH^-

At 5 C, the value of K_w for the equilibrium shown above is 1.9x10^-15 and the value of pK_w is 14.73. Based on this information, which of the following is correct for pure water at this temperature?

• $$\left[H_3O^+\right]=\sqrt[]{1.9\times10^{-15}}$$

• $$pH\ =\ -\log\left(1.9\times10^{-15}\right)$$

• $$14.73\ =\ \left[H_3O^+\right]_{eq}\left[OH^-\right]_{eq}$$

• $$pOH=pH\ +\ 14.73$$

Multiple Choice

In pure water, some of the molecules ionize according to the equation H₂O <--> H⁺ + OH^-. The extent of the ionization increases with temperature. A student heats pure water and records the measured pH at 50 C as 6.6. Based on this information, which of the following mathematical relationships gives the pOH of pure water at 50 C?

• pOH = pH

• $$pOH=\frac{1}{pH}$$

• pOH = 14-pH

• $$pOH\ =\frac{1\times10^{14}}{pH}$$