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Zorluk: KolaypH and pOH Scale and Calculations

Fill in the missing numerical values in the statement below regarding the pH and pOH scale at 25∘C25^\circ\text{C}.

Cevap:For an aqueous solution at 25∘C25^\circ\text{C}, a hydrogen ion concentration of 1.0×10−4 mol dm−31.0 \times 10^{-4}\text{ mol dm}^{-3} corresponds to a pH of 【4】 and a pOH of 【10】.

Cevap

The pH of the solution is 4 (or 4.0) and the pOH is 10 (or 10.0).
Taking the negative logarithm of [H+]=1.0×10−4 mol dm−3[H^+] = 1.0 \times 10^{-4}\text{ mol dm}^{-3} gives a pH of 4. Since pH+pOH=14\text{pH} + \text{pOH} = 14 at 25∘C25^\circ\text{C}, subtracting 4 from 14 gives a pOH of 10.

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1
Calculate the pH from the given hydrogen ion concentration [H+][H^+].
pH=−log⁡10(1.0×10−4)=4\text{pH} = -\log_{10}(1.0 \times 10^{-4}) = 4
pH is defined as the negative logarithm to base 10 of the hydrogen ion concentration.
2
Calculate the pOH using the relationship pH+pOH=14\text{pH} + \text{pOH} = 14.
pOH=14−4=10\text{pOH} = 14 - 4 = 10
At 25∘C25^\circ\text{C}, the sum of pH and pOH for an aqueous solution equals 14.

Anahtar Kavram

Logarithmic pH and pOH calculations from hydrogen ion concentration
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Konular

pH and pOH Scale and Calculations

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