A glass pH electrode behaves linearly with an output change of \(60\) mV per unit change in pH. The measurement reference is set such that a solution of pH \(6\) produces an output of \(60\) mV. The pH of a solution that results in an output of \(-90\) mV is ______.
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The output falls by \(60\) mV for every one unit rise in pH from the reference point. Work out how many pH units correspond to a \(150\) mV drop from \(60\) mV to \(-90\) mV.
Step 1: Note what the calibration tells us.
The electrode reads $60$ mV at $pH = 6$, and every extra unit of pH changes the reading by $60$ mV. We need the pH that gives $-90$ mV.
Step 2: Find how far the output has moved from the reference.
The output has changed from $60$ mV to $-90$ mV, a drop of
\[ \Delta V = 60 - (-90) = 150 \text{ mV} \]
Step 3: Convert the voltage change into a pH change.
Since the electrode output falls by $60$ mV for every one unit rise in pH, a drop of $150$ mV corresponds to a rise in pH of
\[ \Delta pH = \frac{150}{60} = 2.5 \]
Step 4: Add this change to the reference pH.
\[ pH = pH_{ref} + \Delta pH = 6 + 2.5 = 8.5 \]
Step 5: Check the answer.
At $pH=8.5$, the line predicts an output of $60 - 60(8.5-6) = 60 - 150 = -90$ mV, which matches the given reading, so the value checks out.
The other three choices fail this check: putting $pH=3.5$ or $pH=3$ back into the line gives a positive output, not $-90$ mV, and $pH=8$ gives $-60$ mV, not $-90$ mV.
\[ \boxed{pH = 8.5} \]
