Complete this table of values for four aqueous solutions at 25 °C.
Solution A: [H+]=0.0093M [OH-]=? pH=? pOH=?
Solution B: [H+]=?[OH-]=1.4e-6M pH=? pOH=?
Solution C: [H+]=?[OH-]=? pH=5.36 pOH=?
Solution D: [H+]=?[OH-]=? pH=? pOH=10.14
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Each value below represents a different aqueous solution at 25 °C. Classify each solution as acidic, basic, or neutral. pH=13.99 pH=1.77 pOH=8.16 [H+]=4.0e-2 [H+]=5.2e-8 [H+]=1e-7 [OH-]=6.5e-12 [OH-]=8.3e-2 pOH=4.62 pOH=7
To complete the table of values for the four aqueous solutions, you need to understand the relationship between the hydrogen ion concentration [H+], hydroxide ion concentration [OH-], pH, and pOH in an aqueous solution.
1. Solution A:
Given: [H+]=0.0093 M.
To calculate [OH-], use the equation Kw = [H+][OH-] = 1.0x10^-14 M^2 (at 25°C).
Using the given [H+], we can rearrange this equation to solve for [OH-]:
[OH-] = Kw / [H+] = (1.0x10^-14 M^2) / 0.0093 M = 1.08x10^-12 M.
To calculate pH, use the equation pH = -log[H+]:
pH = -log(0.0093) ≈ 2.03.
To calculate pOH, use the equation pOH = -log[OH-]:
pOH = -log(1.08x10^-12) ≈ 11.97.
2. Solution B:
Given: [OH-] = 1.4x10^-6 M.
Use the equation Kw = [H+][OH-] = 1.0x10^-14 M^2 to calculate [H+]:
[H+] = Kw / [OH-] = (1.0x10^-14 M^2) / (1.4x10^-6 M) ≈ 7.14x10^-9 M.
To calculate pH, use the equation pH = -log[H+]:
pH = -log(7.14x10^-9) ≈ 8.15.
To calculate pOH, use the equation pOH = -log[OH-]:
pOH = -log(1.4x10^-6) ≈ 5.85.
3. Solution C:
Given: pH = 5.36.
To calculate [H+], we need to take the antilog of the negative pH value:
[H+] = antilog(-pH) = 10^-pH = 10^-5.36 ≈ 2.28x10^-6 M.
To calculate pOH, we can use the equation pOH = 14 - pH:
pOH = 14 - 5.36 ≈ 8.64.
To calculate [OH-], use the equation Kw = [H+][OH-]:
[OH-] = Kw/[H+] = (1.0x10^-14 M^2) / (2.28x10^-6 M) ≈ 4.39x10^-9 M.
4. Solution D:
Given: pOH = 10.14.
To calculate [OH-], we need to take the antilog of the negative pOH value:
[OH-] = antilog(-pOH) = 10^-pOH = 10^-10.14 ≈ 7.18x10^-11 M.
To calculate pH, we can use the equation pH = 14 - pOH:
pH = 14 - 10.14 ≈ 3.86.
To calculate [H+], use the equation Kw = [H+][OH-]:
[H+] = Kw/[OH-] = (1.0x10^-14 M^2) / (7.18x10^-11 M) ≈ 1.39x10^-4 M.
So, the completed table of values for the four aqueous solutions at 25 °C is:
Solution A: [H+]=0.0093M, [OH-]=1.08x10^-12M, pH=2.03, pOH=11.97
Solution B: [H+]=7.14x10^-9M, [OH-]=1.4x10^-6M, pH=8.15, pOH=5.85
Solution C: [H+]=2.28x10^-6M, [OH-]=4.39x10^-9M, pH=5.36, pOH=8.64
Solution D: [H+]=1.39x10^-4M, [OH-]=7.18x10^-11M, pH=3.86, pOH=10.14.