This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Welcome back BJ(MBM) — missed you this week. Here's the solution to question c) i) and ii): c) The sodium ion level in a urine specimen was measured using an ion-selective electrode. The following values were obtained: 102, 97, 99, 98, 101, 106 mM. The value of t_s for calculating the 95% confidence limit is 2.57 and 99% confidence limits is 4.03. i) Calculate the mean and standard deviation of these values [2.5 marks] ii) What are the 95% and 99% confidence limits for the sodium ion concentration? [2.5 marks] The given data points are x_i = \102, 97, 99, 98, 101, 106\ mM. The number of data points is n = 6. Step 1: Calculate the mean (x). The mean is the sum of all values divided by the number of values. x = ( x_i)/(n) x = (102 + 97 + 99 + 98 + 101 + 106)/(6) x = (603)/(6) x = 100.5 mM Step 2: Calculate the standard deviation (s). The formula for the sample standard deviation is: s = sqrt( (x_i - x))^2n-1 First, calculate (x_i - x)^2 for each data point: • (102 - 100.5)^2 = (1.5)^2 = 2.25 • (97 - 100.5)^2 = (-3.5)^2 = 12.25 • (99 - 100.5)^2 = (-1.5)^2 = 2.25 • (98 - 100.5)^2 = (-2.5)^2 = 6.25 • (101 - 100.5)^2 = (0.5)^2 = 0.25 • (106 - 100.5)^2 = (5.5)^2 = 30.25 Now, sum these values: (x_i - x)^2 = 2.25 + 12.25 + 2.25 + 6.25 + 0.25 + 30.25 = 53.5 Substitute into the standard deviation formula: s = sqrt((53.5)/(6-1)) = sqrt((53.5)/(5)) = sqrt(10.7) s ≈ 3.271 mM Step 3: Calculate the 95% confidence limit. The formula for the confidence limit (CL) is: CL = x ± (t_s · s)/(sqrt(n)) For 95% confidence, t_s = 2.57. CL_95\% = 100.5 ± (2.57 × 3.271)/(sqrt(6)) CL_95\% = 100.5 ± (8.41547)/(2.44949) CL_95\% = 100.5 ± 3.435 CL_95\% = (100.5 - 3.435, 100.5 + 3.435) CL_95\% = (97.065, 103.935) mM Step 4: Calculate the 99% confidence limit. For 99% confidence, t_s = 4.03. CL_99\% = 100.5 ± (4.03 × 3.271)/(sqrt(6)) CL_99\% = 100.5 ± (13.17813)/(2.44949) CL_99\% = 100.5 ± 5.380 CL_99\% = (100.5 - 5.380, 100.5 + 5.380) CL_99\% = (95.120, 105.880) mM i) Mean and Standard Deviation: Mean (x): 100.5 mM Standard deviation (s): 3.271 mM ii) 95% and 99% Confidence Limits: 95% Confidence Limit: (97.065, 103.935) mM 99% Confidence Limit: (95.120, 105.880) mM Drop the next question.