Solve ph and poh questions without having to use log table

Chemistry

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Answer

4
2
11

Step-by-step solution

9 steps

To solve pH and pOH questions without a log table, we typically work with concentrations that are exact powers of 10. This simplifies the logarithm calculation.

The key formulas are:

  • pH=−log⁡[H+]\text{pH} = -\log[\text{H}^+]
  • pOH=−log⁡[OH−]\text{pOH} = -\log[\text{OH}^-]
  • pH+pOH=14\text{pH} + \text{pOH} = 14 (at 25∘C25^\circ\text{C})
  • [H+][OH−]=1.0×10−14[\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} (at 25∘C25^\circ\text{C})

When the concentration of H+\text{H}^+ or OH−\text{OH}^- ions is in the form 1.0×10−x1.0 \times 10^{-x} M, then log⁡(1.0×10−x)=−x\log(1.0 \times 10^{-x}) = -x.

Example 1: Calculating pH from [H+][\text{H}^+]

What is the pH of a solution where the hydrogen ion concentration, [H+][\text{H}^+], is 1.0×10−41.0 \times 10^{-4} M?

  1. Use the pH formula.

    pH=−log⁡[H+]\text{pH} = -\log[\text{H}^+]

  2. Substitute the given concentration.

    pH=−log⁡(1.0×10−4)\text{pH} = -\log(1.0 \times 10^{-4})

  3. Simplify the logarithm.

    Since log⁡(10−4)=−4\log(10^{-4}) = -4.

    pH=−(−4)\text{pH} = -(-4) pH=4\text{pH} = 4

    The pH of the solution is 4\boxed{4}.

    Example 2: Calculating pOH from [OH−][\text{OH}^-]

    What is the pOH of a solution where the hydroxide ion concentration, [OH−][\text{OH}^-], is 1.0×10−21.0 \times 10^{-2} M?

  4. Use the pOH formula.

    pOH=−log⁡[OH−]\text{pOH} = -\log[\text{OH}^-]

  5. Substitute the given concentration.

    pOH=−log⁡(1.0×10−2)\text{pOH} = -\log(1.0 \times 10^{-2})

  6. Simplify the logarithm.

    Since log⁡(10−2)=−2\log(10^{-2}) = -2.

    pOH=−(−2)\text{pOH} = -(-2) pOH=2\text{pOH} = 2

    The pOH of the solution is 2\boxed{2}.

    Example 3: Calculating pH from pOH (or vice versa)

    If the pOH of a solution is 3, what is its pH?

  7. Use the relationship between pH and pOH.

    pH+pOH=14\text{pH} + \text{pOH} = 14

  8. Substitute the given pOH value.

    pH+3=14\text{pH} + 3 = 14

  9. Solve for pH.

    pH=14−3\text{pH} = 14 - 3 pH=11\text{pH} = 11

    The pH of the solution is 11\boxed{11}.

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