Question

you can use this periodic table and reference table of acid and base ionization constants during the quiz. however, as stated in the course outline, and in the orientation session, this quiz is not open book. you cannot access course material while you write the quiz. nor can you google your answers, work as a group, use a cheat sheet, etc.
question 1 (1 point)
what is the ph of a 0.22 mol/l solution of ch₃cooh(aq)?
○ 0.70
○ 1.50
○ 2.70
○ 3.00
○ 0.22
question 2 (1 point)
the notation h₃o⁺ (aq) is sometimes simplified to which of the following?
○ h₂o (l)
○ h⁺ (aq)
○ h⁻ (aq)
○ oh⁻ (aq)
question 3 (1 point)
use the following information to answer the next question.

Explanation:

Question 1

Step 1: Recall the Ka of acetic acid

Acetic acid (\(CH_3COOH\)) is a weak acid with \(K_a = 1.8\times10^{-5}\). For a weak acid \(HA\), the dissociation is \(HA
ightleftharpoons H^+ + A^-\), and the concentration of \(H^+\) can be approximated using \([H^+]=\sqrt{K_a\times c}\) when \(K_a\times c \geq 10^{-10}\) and \(c/K_a \geq 100\) (to check if the approximation is valid). Here, \(c = 0.22\space mol/L\), \(K_a = 1.8\times10^{-5}\). \(c/K_a=\frac{0.22}{1.8\times10^{-5}}\approx12222\geq100\), so the approximation is valid.

Step 2: Calculate \([H^+]\)

Using the formula \([H^+]=\sqrt{K_a\times c}\), substitute \(K_a = 1.8\times10^{-5}\) and \(c = 0.22\space mol/L\):

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Step 3: Calculate pH

\(pH = -\log[H^+]\), substitute \([H^+]=1.99\times10^{-3}\):

$$ pH = -\log(1.99\times10^{-3})\approx2.70 $$

The hydronium ion \(H_3O^+(aq)\) is often simplified to \(H^+(aq)\) in chemical notation because the \(H_3O^+\) is formed when a proton (\(H^+\)) associates with a water molecule, and in many acid - base reactions, the \(H_3O^+\) is represented as \(H^+\) for simplicity, especially in contexts where the water molecule of hydration is not the focus. \(H_2O(l)\) is the solvent, \(OH^-(aq)\) is the hydroxide ion, and the other \(H^+\) option is misformatted. So the notation \(H_3O^+(aq)\) is sometimes simplified to \(H^+(aq)\).

Answer:

2.70

Question 2