Titration help calculations help?

• given 0.2038 gram sample of Ni(NH3)5CL2 dissolved in 0.050L of 0.1503M HCL. Calculate how much 0.09901M NaOH would be required to titrate this sample to a endpoint ? so i know that the HCL in excess converts all the Nh3 in the sample into NH4. so i think you would need to find the moles of NH3 in the sample first that minus that form the moles of hcl ? and go on from there... Could someone show me how to do this step by step please i have a written lab exam coming up it would be a big HELP! please and thank you!

• Answer:

  you know what to do .. here is the solution .. Molar mass of Ni(NH3)5Cl2 is 214.7 g/mole so we have .2038 / 214.7 moles of the nickle salt = 0.000949231 moles there are 5 moles of NH3 in each mole of the salt so we have 0.000949231 x 5 moles of NH3 = 0.004746 moles NH3 + HCl = > NH4 + Cl- so moles of HCl that reacted with the NH3 from the salt = 0.004746 moles moles of HCl added = .1503 x .050 = 0.007515 moles so moles of excess HCl = 0.007515 - 0.004746 = 0.002769 moles of HCl NaOH + HCl = NaCl + H2O moles of NaOH at the end point = 0.002769 volume of .09901 M containing these moles = 0.002769 moles / .09901 mole liter^-1 = 0.02797 L or 27.97 mL