Multiple Choice
What is the pH of a M solution at , assuming complete dissociation?
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17. Acid and Base Equilibrium
pH of Strong Acids and Bases
Multiple Choice
HI is a strong acid (Ka = 3.2 × 109). Calculate [H+], [OH−], pH and pOH of a 7.1 × 10−2 M HI solution.
A
[H+] = 7.1 × 10−2 M
[OH−] = 1.4 × 10−13 M
pH = 1.15
pOH = 12.85
[OH−] = 1.4 × 10−13 M
pH = 1.15
pOH = 12.85
B
[H+] = 7.1 × 10−2 M
[OH−] = 1.4 × 10−13 M
pH = 8.50
pOH = 5.50
[OH−] = 1.4 × 10−13 M
pH = 8.50
pOH = 5.50
C
[H+] = 7.1 × 10−2 M
[OH−] = 7.1 × 10−16 M
pH = 1.15
pOH = 15.15
[OH−] = 7.1 × 10−16 M
pH = 1.15
pOH = 15.15
D
[H+] = 7.1 × 10−2 M
[OH−] = 7.1 × 10−12 M
pH = 1.15
pOH = 11.15
[OH−] = 7.1 × 10−12 M
pH = 1.15
pOH = 11.15
Verified step by step guidance1
Identify that HI is a strong acid, meaning it dissociates completely in water. Therefore, the concentration of hydrogen ions [H+] is equal to the initial concentration of HI, which is 7.1 × 10^−2 M.
Use the ion product of water, Kw, which is 1.0 × 10^−14 at 25°C, to find the concentration of hydroxide ions [OH−]. The relationship is given by: [H+][OH−] = Kw. Substitute [H+] = 7.1 × 10^−2 M into the equation to solve for [OH−].
Calculate the pH using the formula: pH = -log[H+]. Substitute the value of [H+] obtained from the first step into this formula to find the pH.
Calculate the pOH using the formula: pOH = -log[OH−]. Substitute the value of [OH−] obtained from the second step into this formula to find the pOH.
Verify the relationship between pH and pOH, which is given by: pH + pOH = 14. Use the values obtained for pH and pOH to check this relationship.
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