Question

Calibration data for a bromide ion-selective electrode (ISE) was collected and recorded below. The potential of the ISE was measured against a saturated calomel electrode (SCE). All solutions were buffered at a pH of 7.56. A linear calibration curve can be constructed from this data as a plot of potential (in mV) vs. pBr. Determine the slope and y-intercept of such a plot of the calibration data given [Br] (M) Potential (mV) 0.300 0.0300 0.00300 96.4 0.000300 155.1 0.0000300 224.8 Number 21.5 s lope- 36.3 Number y- nte rcept- Two unknown solutions were tested for [Br] using the same bromide ISE. The potential reading for unknown 1 was 13.2 mV and the potential reading for unknown 2 was 116.8 mV. Using the calibration curve you constructed above, determine the concentration of Br in each unknown solution Number Unknown 1 [r- Number Unknown 2 Br-

Answer 1

[Br-l, M log [Br-l mV 3.0E-01 -0.5229 21.5 3.0E-02 1.5229 36.3 3.0E-03 -2.5229 96.4 3.0E-04 -3.5229 155.1 3.0E-05 -4.5229 224.8 250 200 y -61.14x -56.029 R2 0.9987 150 100 50 -5.0000-4.5000-4.0000-3.5000-3.0000-25000-20000-15000-1.0000 、-0.5000 0.0000 log [Br-]

In the graph, Y-axis indicate Where, y Y-axis value, depicts potential in mV ix X-axis value, depicts concertation, log [Br-] m slope # That is, according to the trendline (linear regression) equation y = -61.14x -56.029 , obtained from the graph, 1 potential unit (1 Y Y) is equal to #1. Given, potential of unknown 1 = tial and X-axis depi nc ion. The linear regression equation in in form of y mx X" 61.14 (log x) unit on X-axis (concentration) minus 56.029 13.20 mV Putting y13.20 in trendline equation - 13.20log -61.14 x -56.029 Or, 13.20 56.02961.14 (log x) 69.229 --1.132 61.14 1.132 Hence, [Br] in unknown 1-0.0737 M 10 0.0737 #2. Given, potential of unknown 2 Putting y = 1 16.80 in trendline equation- 116.80log -61.14 x 116.80 mV -56.029 Or, 116.80 56.02961.14 (log x) 172.829 61.14 10 log x--2.827 x= -2.827 Hence, [Br] in unknown 20.0015 M x0.0015 M