Exercises
Exercise 1. What is the hydronium ion concentration, hydroxide ion concentration, the pH, and pOH of a solution that is 0.020 M HNO3?
HNO3 is a strong acid, therefore the [H3O+] concentration is 0.020 M.
The [OH–] is:
\(\displaystyle [OH^-]\;=\;\frac{1.0\times 10^{-14}}{0.020\;M}\;=\;\mathbf{5.00\times 10^{-13}\;M}\)
The pH = -log(0.020) = 1.70
The pOH = 14.00 – 1.70 = 12.30
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Exercise 2. An aqueous solution has a pOH of 13.52. What is the hydronium ion concentration? What is the pH of the solution?
The [OH–] = 10-13.52 = 3.0 x 10-14 M
The [H3O+] is equal to:
\(\displaystyle [H_3O^+]\;=\;\frac{1.00\times 10^{-14}}{3.0\times 10^{-14}\;=\;\mathbf{0.33\;M}}\)
The pH = -log(0.33) = 0.48
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Exercise 3. What is the pH and pOH of a solution that is prepared by dissolving 0.86 g HCl in water to give 250. mL of solution?
First find the molarity of HCl. Convert g to moles. The molar mass of HCl is 36.548 g/mol.
\(\displaystyle 0.86\;g\;HCl\times\frac{1\;mol\;HCl}{36.548\;g\;HCl}\;=\;0.0235\;mol\;HCl\)
Next, calculate the molarity. The volume is 0.250 L.
\(\displaystyle \frac{0.0235\;mole}{0.250\;L}\;=\;0.094\;M\)
HCl is a strong acid therefore the [H3O+] concentration is equal to 0.094 M.
The pH = -log[H3O+] = -log(0.094) = 1.03
The pOH = 14.00 – 1.03 = 12.97
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Exercise 4. Calculate the hydronium ion concentration, the hydroxide ion concentration, the pH, and the pOH by diluting 45.0 mL of 0.68 M Ba(OH)2 to a volume of 250.0 mL?
First find the molarity of the diluted Ba(OH)2
moles Ba(OH)2 = 0.045 L x 0.68 M = 0.0306 moles Ba(OH)2
Find the molarity by dividing by the volume in L.
\(\displaystyle M\;=\frac{0.0306\;mol}{0.250\;L}\;=\;0.122\;M\)
There are two moles of OH– in one mole of Ba(OH)2.
The solution is 2 x 0.122 M = 0.24 M OH–
The [H3O+] = 1.0 x 10-14/0.24 M = 4.17 x 10-14
The pH = -log(4.17 x 10-14) = 13.38
The pOH = 14.00 – 13.38 = 0.62
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Exercise 5. How many grams of CaO should be dissolved in water to make 1.50 L of a solution with a pH of 11.20?
First, calculate the hydroxide ion concentration. The oxide ion will react with hydrogen ion to form OH– ion.
First calculate the hydronium ion concentration.
[H3O+] = 10-pH = 10-11.20 = 6.31 x 10-12 M.
Calculate the hydroxide ion concentration.
\(\displaystyle [OH^-]\;=\;\frac{K_w}{[H_3O^+]}\;=\;\frac{1.0\times 10{-14}}{6.31\times 10^{-12}M}\;=\;1.6\times 10^{-3}\;M\)
Next we use the chemical equation to determine the number of grams of CaO required.
CaO (s) + H2 (l) → Ca2+ (aq) + 2 OH– (aq)
\(\displaystyle 1.6\times 10^{-3}\;mol\;OH^-\times\frac{1\;mol\;CaO}{2\;mol\;OH^-}\times\frac{56.0774\;g\;CaO}{1\;mol\;CaO}\;=\;\mathbf{0.044\;g\;CaO}\)