Question

Search Textb In The Labor Review Topicsl References Use the References to access important values if needed for this question. In the laboratory a student combines 36.4 ml. of a 0.111 M potassium nitrate solution with 12.6 mL of a 0.544 M potassium phosphate solution. What is the final concentration of potassium cation ? 0.222 M Submit Answer Retry Entire Group 6 more group attempts remaining

Answer 1

Answer Number of mole of Potassium nitrate (KNO3). ole of KNO3 = volume of KNO3 (en L x molarity of KNO3 Solution 36.4 LX0-111 Mol 1ooo = 0.00404 mol 4.0404 looo KNO3 ug) - ktor) + NO3(ar) 0.00404 mol 0.00404 mol 0.00404 mol (2) plumber of mole of Potassiumm Phosphate (K3 P04) molarity of mole af Kap oa = Volume of K3 P04 (ebL) x K3 PO 4 Solution = 12-6 tx 0.544 mol 1000 ㅂ = 6.8544 = 0.00685 mol K3 PO4 (aa) 0.00685 mol poter) - 3 kt (ar) + 38 0.00685 ²0.02056moll 0.00685 mol Total Number of mole ces Kt cation - 0.00404 + 0.02056 – 0.0246 mol Total Wolume of Solution - 36.4 t 12.6 mL. 49.0 me

Concent volume of me ekto L. 0.049L = Total volume 49.0 = - Number of mole of Kt cation molarity of Kt cation Tototal voleone of solation in L). 0.0246 mol 6.0492 0.502 mol Final concentration of Krcetion = 0.502 M "Pleale rakit "