Question

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wednesday - sem 1 - toronto dsb

question 15 (1 point)
what is the ph if the concentration of oh⁻ = 3.1 × 10⁻¹¹ mol/l?
○ 11
○ 10.5
○ 3.49
○ 2.51

question 16 (1 point)
how are arrhenius acids and brønsted-lowry acids similar?
○ both produce hydroxides
○ both produce protons
○ both are bonded to hydrogen
○ both are bonded to hydroxides

question 17 (1 point)
an acid has a percentage ionization of 3.3 %. if it has an initial concentration of 2.5 mol/l, then what is the change in concentration? (hint: determine the value of c in the ice table.)
○ 2.5 mol/l, since x (from the 100 rule) is so small
○ 0.08 mol/l
○ 2.42 mol/l
○ 8.25 mol/l

question 18 (1 point)
solve for ka if kb = 1 × 10⁻⁵

Explanation:

Question 15

Step 1: Find \( H^+ \) from \( OH^- \)

We know that \( K_w = [H^+][OH^-] = 1.0\times10^{-14} \) at 25°C. Given \( [OH^-] = 3.1\times10^{-11} \, \text{mol/L} \), we can find \( [H^+] \) as:

$$ [H^+] = \frac{K_w}{[OH^-]} = \frac{1.0\times10^{-14}}{3.1\times10^{-11}} $$

$$ [H^+] \approx 3.2258\times10^{-4} \, \text{mol/L} $$

Step 2: Calculate pH

The formula for pH is \( pH = -\log[H^+] \). Substituting \( [H^+] \):

$$ pH = -\log(3.2258\times10^{-4}) $$

$$ pH \approx 3.49 $$

Arrhenius acids produce \( H^+ \) (protons) in aqueous solution. Brønsted - Lowry acids are proton donors. So both produce protons. The option "both produce hydroxides" is wrong as acids don't produce hydroxides. "both are bonded to hydroxides" is wrong. "both are bonded to hydrogen" is not the defining similarity of their acidic behavior.

Step 1: Recall the formula for percentage ionization

Percentage ionization \(=\frac{\text{Change in concentration}(x)}{\text{Initial concentration}(c)}\times100\)
Given percentage ionization \( = 3.3\%=0.033\) and initial concentration \( c = 2.5\,\text{mol/L}\)

Step 2: Solve for \( x \)

From the formula \( 0.033=\frac{x}{2.5}\times100 \)

$$ x=\frac{0.033\times2.5}{100}\times100? \text{Wait, correct formula: } x = c\times\frac{\text{percentage ionization}}{100} $$

$$ x=2.5\times\frac{3.3}{100}=2.5\times0.033 = 0.0825\approx0.08\,\text{mol/L} $$

Answer:

3.49

Question 16