Question

Potential (mV) [Br](M) 0.300 Calibration data for a bromide ion-selective electrode (ISE) was collected and recorded in the table. 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. -21.2 0.0300 36.2 95.7 0.00300 0.000300 155.5 0.0000300 225.1 slope = mVy-intercept = mV Two unknown solutions were tested for [Br] using the same bromide ISE. The potential reading for unknown 1 was 27.8 mV, and the potential reading for unknown 2 was 155.3 m V. Using the calibration curve you constructed, determine the concentration of Br in each unknown solution. unknown 1 [Br] = M unknown 2 [Br] =

Answer 1

slope = - 61.19 M

1 0.3 -21.2 -0.5228 -21.2 2 0.03 36.2 -1.5228 36.2 250 3 0.003 95.7 -2.5228 95.7 4 0.0003 155.5 -3.5228 155.5 y -61.19x-56.11 R2 0.998 200 0.00003 5 225.1 -4.5228 225.1 6 [Br- mV pBr mV 150 7 potental (mV) 100 8 9 50 10 11 12 -5 0 50 -4 -3 -2 1 13 14 15 -100 pBr 16 17

Intercept = - 56.11 mV.

Now the bestline equation is

Y = - 61.19 x - 56.11

Y is potential.

X is pBr

For unknown 1

Now, when, y = 27.8 mv

Then

27.8 = -61.19× pBr -56.11

Or, - 61.19 pBr = 83.91

Or, pBr = - 1.371

Or, [Br^-] = 10^-1.371 = 0.0426 M

Unknown 2.

again , when potential = 155.3 mV.

155.3 = -61.19 × pBr - 56.11

Or, -61.19 pBr = 211.41

Or, pBr = - (211.41/61.19)

Or, pBr = - 3.45

Or, [Br^-] = 10^(-0.345)

Or, [Br^-] = 0.000355 M

{ pBr = - log [Br^-] }.

Then, [Br^-] = 10^-(pBr) .