A sample of orange juice is found to have a PH of 3.80. What is the concentration of the hydroxide ion in the juice?

Analysis

To find the concentration of the hydroxide ion $[OH^-]$ in a solution when given the pH, we must use the relationship between pH, pOH, and the ion product constant of water ($K_w$).

Concept

1. pH and pOH relationship: At $25^\circ C$, the sum of pH and pOH is always 14. • $pH + pOH = 14$
2. Calculating pOH: Once you have the pH, you find the pOH by subtracting the pH from 14. • $pOH = 14 - pH$
3. Calculating $[OH^-]$: The concentration of hydroxide ions is calculated using the formula: • $[OH^-] = 10^{-pOH}$

Step-by-Step Solution

Step 1: Calculate the pOH. Given $pH = 3.80$ $pOH = 14 - 3.80$ $pOH = 10.20$

Step 2: Calculate the hydroxide ion concentration $[OH^-]$. $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{-10.20}$

To solve $10^{-10.20}$ without a calculator: $10^{-10.20} = 10^{0.80} \times 10^{-11}$ Using the value $10^{0.80} \approx 6.31$ (since $\log_{10}(6.31) \approx 0.80$)

Therefore: $[OH^-] \approx 6.3 \times 10^{-11} \, mol/dm^3$

Conclusion

Comparing our result to the given options, the calculated value matches option B.

Exam Tip

In JAMB Chemistry, always remember the "14 rule." If you are given pH and asked for $[OH^-]$, you must convert to pOH first. If you are given pOH and asked for $[H^+]$, convert to pH first. Also, keep in mind that $10^{-10.2}$ is the same as $10^{0.8} \times 10^{-11}$; knowing your basic log values (like $\log 2 \approx 0.3$, $\log 3 \approx 0.47$, $\log 6 \approx 0.78$) helps you estimate these answers quickly without a calculator.