Question

A 23.28mL solution containing 1.684 g Mg(NO3)2 is mixed with a 31.67mL solution containing 1.172 g NaOH. Calculate the concentrations of the Na + and NO3- ions that remain in solution. Assume volumes are additive. Balanced equation not needed to solve the problem.

Answer 1

We know that concentration - h ne nuber of moles va volume. As we mixed the best solutions Hence the final volume WHV= 23.28 + 81067 = 54.95 mL Now the number of moles Maro present in 23.28 mL of 1.684g Soletionis n = W = Amount of salt

20 aber - n = W = Amount of salt 21.684g . = Molecular weight of salt = 148.3 g/mol in e na 1.684 g o m 148.3 g/mol ou danas 0.0011 mol regeorde) ? reg 2 + 4 2005 2 p Again 1 I mole rag(NO) Conturs 2 mole No ion . So number of mole of not z 2x0.011 = 0.022 mole concentration of Noz is solution s o.022 mol 0.022 mot let ur vom 54.95 m 54.95632 mup a 0.4 mol/L 0.4M

22.28 mL of 1.684 9 MO(NO) is Numbers of mole of wooh 31.67 mi solution Containing 1.172 g NaOH is - e l: 172 g - 29.997 Monor Atolecular weight of Naot 39.9 2 0.029 mole Now, Naot Nat toch se 1 mole Naot produce Imole nat So Number of mole of Nat = 0.029 mm Concentration of wat 7.0.029 mor - 0.029 mol 54.95 mL - 54.958103 L = 0.5 mol/L = o.sm