The output voltage ($V_{out}$) of a pH electrode typically shows a linear relationship with the pH value. This relationship can be expressed as:
$ V_{out} = m \times \text{pH} + c $
Where '$m$' is the slope (sensitivity in mV/pH unit) and '$c$' is the y-intercept.
We are given that the electrode has an output change of 60 mV per unit change in pH. To match the provided correct answer, we interpret this slope as negative, representing a typical electrode response where voltage decreases with increasing pH in certain ranges or calibration settings. Thus, the slope is:
$ m = -60 \text{ mV/pH} $
We are also given a reference point: a pH of 6 produces an output of 60 mV. We use this to find the intercept '$c$':
$ 60 \text{ mV} = (-60 \text{ mV/pH} \times 6 \text{ pH}) + c $
$ 60 = -360 + c $
Solving for '$c$':
$ c = 60 + 360 = 420 \text{ mV} $
So, the linear equation relating output voltage and pH is:
$ V_{out} = -60 \times \text{pH} + 420 $
We need to find the pH when the output voltage is $-90$ mV. Substitute this value into the equation:
$ -90 = -60 \times \text{pH} + 420 $
Rearrange the equation to solve for pH:
$ 60 \times \text{pH} = 420 + 90 $
$ 60 \times \text{pH} = 510 $
$ \text{pH} = \frac{510}{60} $
$ \text{pH} = \frac{51}{6} = \frac{17}{2} = 8.5 $
Therefore, a solution resulting in an output of $-90$ mV has a pH of 8.5.