Question

A 0.4528 g sample of a pure carbonate, XnCO3(s) was dissolved in 50.0 mL of 0.1400 M HCl(aq). The excess HCl(aq) was back titrated with 24.60 mL of 0.0980 M NaOH(aq).

How many moles of HCl react with the carbonate?

What is the identity of the cation, X?

Answer 1

total no of moles of HCl   = molarity * volume in L

                                        = 0.14*0.05   = 0.007 moles

no of moles of NaOH = molarity * volume in L

                                   = 0.098*0.0246   = 0.0024 moles

HCl + NaOH ------------------> NaCl + H2O

1 mole of NaOH react with 1 mole of HCl

0.0024 moles of NaOH react with 0.0024 moles of HCl

no of moles of excess HCl   = total no of moles of HCl - no of moles of HCl

                                              = 0.007 -0.0024   = 0.0046 moles of excess HCl

XnCO3(s) + 2HCl -----------------------> XnCl2 + Co2 + H2O

2 moles of HCl react with 1 mole of XnCO3

0.0046 moles of HCl react with = 1*0.0046/2   = 0.0023 moles of XnCO3

molar mass of XnCO3 = mass of XnCO3/no of moles of XnCo3

                                      = 0.4528/0.0023   = 197g/mole

molar mass of Co3^2-    = 12+16*3   = 60g/mole

molar mass of Xn^2+               = molar mass of XnCO3 - molar mass of Co3^2-

                                               = 197-60   = 137g/mole

That is Ba^2+

the identity of the cation, X Ba >>>>>>>>>>>>>answer