Question

A mixture was made from solutions of sodium sulfate and aluminium sulfate. 313 mL of 0.500 M Na2SO4 solution and 187 mL of 0.300 M Al2(SO )3 solution were mixed together. Calculate the concentration of SO 2-ions in the final solution?

Answer 1

kindly post different questions separately. please rate

mmol ( milimol) = concentration (M) X volume (mL) mmul a NoaS0y = 0.500 M X 313 mL = 156.5 mmol mmold Ala (904)3 = 0.300 M X 187m2 = 56.1 mmol Nqs804 dissociates completely in aqueous solution Nassou (aq) - 2 Nat lag) + suya-lag) Aftex dmoctation 2 x 156.5 mmol 156.5 mmol = 313 mmol mmol ol s042from N92804 = 156.5 mmol = 0 Alacsvydz dissociates completely in aqueous solution Al 1904/3 (aq) - 2013+(aq) + 35042-199) After dimoctation « X 56.) mmol 3x:56.1 mmol - 112,2 mmol = 168.3 mmol mmol of soy2 from Ala (304/3 = 168.3 mmol - ② After adding both solution final volume of solution - from 1 and 313 ml + 187 mL = 500mL Total mmol of Soy2 = (156.5 + 168.3) mmol = 324.8 mmol [s042-] = mmol Total volume (mel 324,8 mmol = 0.6496 M 500 m2 Hence, the concentration is 0.6496 M of soy2- ions in the final solution