Examkin

Question

Difficulty: EasypH 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}.

Answer: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】.

Answer

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.

Step-by-Step Solution

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.

Key Concept

Logarithmic pH and pOH calculations from hydrogen ion concentration
Rate this question

Topics

pH and pOH Scale and Calculations

More questions from this topic