Question

Module 22. Question 17. Homework. Unanswered An electrochemical cell using a calcium ion ISE is found to exhibit the calibration relationship of E (V) = 0.208 - (0.0283 *pCa). If the ISE is placed into a solution and a voltage of 0.128 V is recorded, what is the concentration of calcium ion present? Numeric Answer:

Answer 1

Given. ELV) = 0.208 - (0.0283X pla] ELV)= 0.128 V . concentration of ca = ? 0.128 = 0.208 - 10.0283xpca] 0.0283 *pca = 0.208-0.128 0.02834pca -0.08 _pla= 0.08 0.0283 pca 2 2.8268 Raz pa = -log[ca [ca] = 10 pca -2.8268 - 10 [ca] = 1:49x10² concentration of caleidhm in present = 1.49810 M = 1.49 MM