Question

GENERAL CHEMISTRY STUDY CHEMISTRY STUDY SHEET NO. 16 How many mL of a 0.121 M Ba(OH), solution is required to completely new tralize 22.4 mL of a 0.210 M H.PO, solution? 2. Tt requires 27.5 mL of a NaOH solution to titrate 0.214 g of H.C.O. (MM = 90.01 molarity of the NaOH solution? 3. It requires 42.5 mL of a NaOH solution to titrate 25.0 mL of a 0.192 M HCL evaluation TE 92 m of the M OT 1

can someone explain to me how #1 is solved. please dont just solve the problem, explain the steps to get the answer. I have trouble understanding the concept

Answer 1

0.21 M H3PO4 Solution means - 1000 mL 22.4mL H₂POg solution contains 0.21 mole H₂PO4 H₃PO4 son contains 0.21 - x 22.4) mole 1000-4-1 more H₂POg = 4078103 mole H₂PO2- Reaction : 2 H₂PO4 + 3 Ba(OH)2 Ba, (PO4)₂ + 6 H₂O 2 mole H3 PO4 Yeacts with 3 mole Ba(OH), i 4.78103 mole H3POq will react with 3 x 4.7x103) = 7.05X163 mole Ba(OH) Lets x ml of 0.121 M Ba(OH), contains 7.05x103 mole BacOH)as it reacts completely with the aid. similarly, 0,121 M BaOH), solution means - 1000 mL Ba(OH), soin contains 0.121 mote Balon ;. x mL 99 99 99 ( 0.121 xx) mole Ba(OH)2 = 1.21 x 109 x more. NOW, 1.21x109 x x = 7.058103 . x = 58.26

Hence, 58.26 ml of 0.121 M BaOH), son is required to completely neutralize 22.4mL of 0.21 M H₂PO2. Reaction: H₂204 + 2 Naot Naz 20 + 2 H2O. Molar mass of H₂CO = 90 g/mol. We know, no. of moles = given mass Molar mass 0.214 g 0.214 g H₂ C2O4 = H₂C2O4 90 g/mol = 2.38×10 3 mol H₂C2Og. from the reaction stoichiometry we can say that 2-38 x 103 mole H₂C2O4 will react with 2X 2.38x153 mole Nach = 4.76 x 103 mole NaOH. Hene, 27.5 mL Noot contains 4.76 x 103 mole Naoh. NaOH 2018 will contain (4.768103 1000) -X10002 27.5 i 1000 m = 0.173 mole Naoh Naot soin is 0.173M. Hence, molarity of the