Question

A standard solution of ammonia is 100.0 ppm. You use an ion selective electrode (ISE) to measure ammonia of this standard solution and obtain the following data (in ppm): 99.4, 101.1, 100.4, 98.0, 99.6. What is the standard deviation of the mean of ammonia (in ppm) determined by the ISE? Report your answer to two significant figures.

I got 1.2 and that's incorrect.

Answer #1

Solution :

n = 5

= = 498.5 / 5 = 99.7

s = (x
- )^{2} / n -
1

= (99.4 -
99.7)^{2} + (101.1 - 99.7)^{2} + (100.4 -
99.7)^{2} + (98 - 99.7)^{2} + (99.6 -
99.7)^{2} / 5 - 1

= 5.44 / 4

**= 1.17**

The calcium ion selective electrode (ISE) measures 11.8 mV in a
1 mM Ca2+ solution at room temperature. Then the ISE is
placed in another solution at body temperature and it measures 20
mV. What is the calcium concentration of the new solution? To avoid
complicated theory assume that Eo is not Temperature
dependent.

Fluoride ion selective electrodes obey the following
equation:
E(V) = K -
β(0.0592)log[F-]
where the fluoride concentration is expressed as mol/L and the
temperature coefficient β is normally equal to 1.
In a normal calibration procedure you dilute a standard fluoride
solution to between 10 and 20 mg/L fluoride ion and then measure
the response of the electrode. You then plot the data and do a
linear regression to obtain the fitting coefficients K and β. Use
the plotting and...

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