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
All of these follow two/three formulas.
(H+)(OH^-) = Kw = 1E-14
pH = -log(H^+)
pOH = -log(OH^-)
pH + pOH = pKw = 14.
1*10
To complete the table of values for the aqueous solutions A, B, C, and D, we will use the following relationships:
1. pH + pOH = 14
2. [H+] × [OH-] = 1.0 × 10^-14
Let's start with Solution A:
Solution A:
Given: [H+] = 0.0093 M
We can calculate [OH-] using the second relationship:
[OH-] = 1.0 × 10^-14 / [H+]
[OH-] = 1.0 × 10^-14 / 0.0093
[OH-] ≈ 1.08 × 10^-12 M
To calculate the pH, we can use the formula:
pH = -log10[H+]
pH = -log10(0.0093)
pH ≈ 2.03
Finally, we can calculate the pOH using the first relationship:
pOH = 14 - pH
pOH = 14 - 2.03
pOH ≈ 11.97
Solution A: [H+] = 0.0093 M, [OH-] ≈ 1.08 × 10^-12 M, pH ≈ 2.03, pOH ≈ 11.97
Now let's move on to Solution B:
Solution B:
Given: [OH-] = 1.4 × 10^-6 M
Using the second relationship, we can find [H+]:
[H+] = 1.0 × 10^-14 / [OH-]
[H+] = 1.0 × 10^-14 / 1.4 × 10^-6
[H+] ≈ 7.14 × 10^-9 M
To find the pH, we can apply the formula:
pH = -log10[H+]
pH = -log10(7.14 × 10^-9)
pH ≈ 8.15
From here, we can calculate the pOH:
pOH = 14 - pH
pOH = 14 - 8.15
pOH ≈ 5.85
Solution B: [H+] ≈ 7.14 × 10^-9 M, [OH-] = 1.4 × 10^-6 M, pH ≈ 8.15, pOH ≈ 5.85
Moving on to Solution C:
Solution C:
Given: pH = 5.36
To find [OH-], we need to find [H+] using the inverse of the pH:
[H+] = 10^-pH
[H+] = 10^-5.36
[H+] ≈ 2.51 × 10^-6 M
Using the second relationship, we can calculate [OH-]:
[OH-] = 1.0 × 10^-14 / [H+]
[OH-] = 1.0 × 10^-14 / 2.51 × 10^-6
[OH-] ≈ 3.98 × 10^-9 M
Lastly, we can calculate the pOH:
pOH = 14 - pH
pOH = 14 - 5.36
pOH ≈ 8.64
Solution C: [H+] ≈ 2.51 × 10^-6 M, [OH-] ≈ 3.98 × 10^-9 M, pH = 5.36, pOH ≈ 8.64
Lastly, let's calculate Solution D:
Solution D:
Given: pOH = 10.14
To find [OH-], we can use the inverse of pOH:
[OH-] = 10^-pOH
[OH-] = 10^-10.14
[OH-] ≈ 7.72 × 10^-11 M
Using the second relationship, we can calculate [H+]:
[H+] = 1.0 × 10^-14 / [OH-]
[H+] = 1.0 × 10^-14 / 7.72 × 10^-11
[H+] ≈ 1.30 × 10^-4 M
Finally, we can calculate the pH:
pH = -log10[H+]
pH = -log10(1.30 × 10^-4)
pH ≈ 3.89
Solution D: [H+] ≈ 1.30 × 10^-4 M, [OH-] ≈ 7.72 × 10^-11 M, pH ≈ 3.89, pOH = 10.14
Therefore, the completed table of values for the four aqueous solutions at 25°C is as follows:
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| Solution | [OH-] (M) | pH | pOH |
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| A | 1.08 × 10^-12 | 2.03 | 11.97 |
| B | 1.4 × 10^-6 | 8.15 | 5.85 |
| C | 3.98 × 10^-9 | 5.36 | 8.64 |
| D | 7.72 × 10^-11 | 3.89 | 10.14 |
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To complete the table of values for the four aqueous solutions, we need to apply some concepts related to the properties of acids and bases:
1. Definition of pH and pOH:
pH is a measure of the acidity or alkalinity of a solution and is defined as the negative logarithm (base 10) of the hydrogen ion concentration ([H+]) in moles per liter:
pH = -log[H+]
Similarly, pOH is the negative logarithm (base 10) of the hydroxide ion concentration ([OH-]) in moles per liter:
pOH = -log[OH-]
2. Relationship between pH and pOH:
Since water is neutral, the product of [H+] and [OH-] is always 1. That means:
[H+] x [OH-] = 1 x 10^-14
Taking the logarithm of both sides gives:
-log[H+] - log[OH-] = -log(1 x 10^-14)
Thus:
pH + pOH = 14
Now let's analyze each solution step by step:
Solution A:
[H+] = 0.0093M (given)
Since [H+] and [OH-] are related by the Kw expression, we can calculate [OH-]:
[OH-] = Kw / [H+] = 1 x 10^-14 / 0.0093 ≈ 1.07 x 10^-12M
To find pH, we can use the equation pH = -log[H+]:
pH = -log(0.0093) ≈ 2.03
To calculate pOH, we can use the equation pOH = 14 - pH:
pOH = 14 - 2.03 ≈ 11.97
Solution B:
[H+] = ? (given)
[OH-] = 1.4 x 10^-6M (given)
Using the Kw relationship, we can calculate [H+] from [OH-]:
[H+] = Kw / [OH-] = 1 x 10^-14 / 1.4 x 10^-6 ≈ 7.14 x 10^-9M
To find pH, we use pH = -log[H+]:
pH = -log(7.14 x 10^-9) ≈ 8.15
To calculate pOH, we use the equation pOH = 14 - pH:
pOH = 14 - 8.15 ≈ 5.85
Solution C:
[H+] = ? (given)
pH = 5.36 (given)
To find [OH-], we can use the equation [H+] x [OH-] = 1 x 10^-14. Rearranging the equation:
[OH-] = 1 x 10^-14 / [H+]
Substituting the given pH into the equation:
[OH-] = 1 x 10^-14 / 10^-5.36 ≈ 2.48 x 10^-10M
To find pOH, we can use the equation pOH = -log[OH-]:
pOH = -log(2.48 x 10^-10) ≈ 9.61
Solution D:
[H+] = ? (given)
pOH = 10.14 (given)
To find [OH-], we can use the equation pOH = -log[OH-]:
[OH-] = 10^(-pOH) = 10^(-10.14) ≈ 5.2 x 10^(-11)M
To find pH, we use the equation pH + pOH = 14:
pH ≈ 14 - 10.14 ≈ 3.86
Now the completed table of values is as follows:
Solution A: [H+]=0.0093M [OH-]=1.07 x 10^-12M pH=2.03 pOH=11.97
Solution B: [H+]=7.14 x 10^-9M [OH-]=1.4 x 10^-6M pH=8.15 pOH=5.85
Solution C: [H+]=2.48 x 10^-10M [OH-]=1 x 10^-14 / 10^-5.36M pH=5.36 pOH=9.61
Solution D: [H+]=10^-pHM [OH-]=5.2 x 10^-11M pH=3.86 pOH=10.14